Yes. The equality can in fact be reached. Our strategy will be to produce a compact set $K_{d,r}\subseteq M_{d}(\mathbb{C})^{r}$ such that if $(X_1,\dots,X_r)\in K_{d,r}$, then $\rho(\Phi(X_1,\dots,X_r))=1$, and where
$$\rho_{2,d}(A_1,\dots,A_r)=\max\{\rho(A_1\otimes X_1+\dots+A_r\otimes X_r)\mid(X_1,\dots,X_r)\in K_{d,r}\}.$$
Let us go over a few definitions and facts to give some context to our construction. These facts can easily be found in John Watrous's 2018 book called The Theory of Quantum Information.
A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be positive if whenever $X$ is positive semidefinite, $\mathcal{E}(X)$ is also positive semidefinite. A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be completely positive if the mapping $\mathcal{E}\otimes 1_{U}:L(V\otimes U)\rightarrow L(W\otimes U)$ is positive for each finite dimensional complex Hilbert space $U$.
A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be trace preserving if $\text{Tr}(\mathcal{E}(X))=\text{Tr}(X)$ for each $X\in L(V)$.
Proposition: Let $\mathcal{E}:L(V)\rightarrow L(W)$ be a linear operator.
$\mathcal{E}$ is completely positive if and only if there are $A_1,\dots,A_r$ where $\mathcal{E}=\Phi(X_1,\dots,X_r)$.
If $\mathcal{E}$ is defined by letting $\mathcal{E}(X)=A_1XB_1^*+\dots+A_rXB_r^*$, then $\mathcal{E}$ is trace preserving if and only if $A_1^*B_1+\dots+A_r^*B_r=1_V$.
A quantum channel is a completely positive trace preserving operator $\mathcal{E}:L(V)\rightarrow L(W)$. In quantum information theory, the quantum channels are the main morphisms between quantum states.
Proposition: If $\mathcal{E}$ is a quantum channel, then $\rho(\mathcal{E})=1$.
Let $K_{d,r}$ be the collection of all tuples $(X_1,\dots,X_r)\in M_{d}(\mathbb{C})^r$ such that $X_1^*X_1+\dots+X_r^*X_r=1_d$. Said differently, $K_{d,r}$ is the collection of all tuples $(X_1,\dots,X_r)\in M_{d}(\mathbb{C})^r$ such that
$\Phi(X_1,\dots,X_r)$ is a quantum channel. The set $K_{d,r}$ is clearly a closed set, and $K_{d,r}$ is bounded since $d=\text{Tr}(X_1^*X_1+\dots+X_r^*X_r)=\|X_1\|_2^2+\dots+\|X_r\|_2^2$ where $\|\cdot\|_2$ denotes the Frobenius norm, so $K_{d,r}$ is compact.
Lemma: Let $A_1,\dots,A_r\in L(V)$. Suppose that there is no $x\in V\setminus\{0\}$ with $A_1x=\dots=A_rx=0$. Furthermore, suppose that there is no subspace $W\subseteq V$ with $W\neq\{0\},W\neq V$, and $W=A_1[W]+\dots+A_r[W]$. Then there is a $\lambda>0$ along with a positive definite $P$ with $\Phi(A_1,\dots,A_r)(P)=\lambda P$.
Proof: Now, let $\mathcal{Q}$ be the collection of all positive semidefinite matrices in $L(V)$ with trace $1$, and let $F:\mathcal{Q}\rightarrow\mathcal{Q}$ be the mapping defined by letting $$F(P)=\frac{\Phi(A_1,\dots,A_r)(P)}{\text{Tr}\big(\Phi(A_1,\dots,A_r)(P)\big)}.$$ Then $\mathcal{Q}$ is convex, and $F$ is a continuous bijection, so by the Brouwer fixed point theorem, there is some $P\in\mathcal{Q}$ with
$F(P)=P$. Therefore, we have $\Phi(A_1,\dots,A_r)(P)=\lambda P$ for some positive $\lambda$.
Now,
$$\text{Im}(P)=\text{Im}(\lambda P)=\text{Im}(A_1PA_1^*+\dots+A_rPA_r^*)$$
$$=\text{Im}(A_1PA_1^*)+\dots+\text{Im}(A_rPA_r^*)=\text{Im}(A_1P)+\dots+\text{Im}(A_rP)$$
$$=A_1[\text{Im}(P)]+\dots+A_r[\text{Im}(P)].$$
This is only possible if $\text{Im}(P)=V$. $\square$
Let $O_{d,r}$ be the collection of all $(X_1,\dots,X_r)\in M_{d}(\mathbb{C})^{r}$ where $\Phi(X_1,\dots,X_r)$ is not nilpotent. Let $E_{d,r}$ be the collection of all $(X_1,\dots,X_r)\in M_d(\mathbb{C})^r$ where there is no subspace $W\subseteq\mathbb{C}^d$ with
$W=X_1^*[W]+\dots+X_r^*[W]$ and $W\neq\{0\},W\neq\mathbb{C}^d$ and where there is no $x$ with $X_j^*x=0$ for $1\leq j\leq r$.
The sets $O_{d,r},E_{d,r}$ are dense subsets of $M_{d}(\mathbb{C})^{r}$ with $E_{d,r}\subseteq O_{d,r}$.
Proposition: Suppose that $(X_1,\dots,X_r)\in E_{d,r}$. Then there is an invertible matrix $B$ and some positive number $\lambda$ with $(\lambda BX_1B^{-1},\dots,\lambda BX_rB^{-1})\in K_{d,r}$.
Proof: Suppose that $B$ is invertible and $\lambda$ is positive. Then $(\lambda BX_1B^{-1},\dots,\lambda BX_rB^{-1})\in K_{d,r}$ if and only if
$$\sum_{k=1}^{r}\lambda^2 (B^{-1})^*X_k^*B^*BX_k^*B^{-1}=I$$ if and only if
$$\sum_{k=1}^{r}\lambda^2 X_k^*B^*BX_k=B^*B$$.
Therefore, there are $\lambda,B$ with $(\lambda BX_1B^{-1},\dots,\lambda BX_rB^{-1})\in K_{d,r}$ if and only if there is a positive $\mu$ and a positive definite $P$ with $$\mu\Phi(X_1^*,\dots,X_k^*)P=P.$$
On the other hand, the existence of such a positive definite $P$ and positive $\mu$ is guaranteed by the above lemma. $\square$
Now, define a mapping $G_{A_1,\dots,A_r}:O_{d,r}\rightarrow\mathbb{R}$ by letting
$$G_{A_1,\dots,A_r}(X_1,\dots,X_r)=\frac{\rho(A_1\otimes X_1+\dots+A_r\otimes X_r)}{\rho(\Phi(X_1,\dots,X_r))^{1/2}}.$$ Since the mapping $G$ is continuous, we have
$$\rho_{2,d}(A_1,\dots,A_r)=\sup\{G_{A_1,\dots,A_r}(X_1,\dots,X_r)\mid (X_1,\dots,X_r)\in E_{d,r}\}.$$
Since $$G_{A_1,\dots,A_r}(X_1,\dots,X_r)=G_{A_1,\dots,A_r}(\lambda BX_1B^{-1},\dots,\lambda BX_rB^{-1})$$ whenever $B$ is invertible and $\lambda$ is a non-zero complex number, by the above proposition, we know that
$$\{G_{A_1,\dots,A_r}(X_1,\dots,X_r)\mid (X_1,\dots,X_r)\in E_{d,r}\}$$
$$=\{G_{A_1,\dots,A_r}(X_1,\dots,X_r)\mid (X_1,\dots,X_r)\in K_{d,r}\}.$$
Therefore, since $K_{d,r}$ is compact, there is some $(Z_1,\dots,Z_r)\in K_{d,r}$ with
$$G_{A_1,\dots,A_r}(Z_1,\dots,Z_r)=\max\{G_{A_1,\dots,A_r}(X_1,\dots,X_r)\mid (X_1,\dots,X_r)\in K_{d,r}\}$$
$$=\rho_{2,d}(A_1,\dots,A_r).$$
I have ran computations that maximize $G_{A_1,\dots,A_r}$, and in these computations the maximum seems to actually be reached.
The answer to all these questions is Yes. We shall answer this question in the more general case when $X_1,\dots,X_r$ are complex matrices (the inequality for this question looks a little bit different in the complex case since we need to apply conjugations and transposes). This answer shall be in the context of quantum channels. I recommend The Theory of Quantum Information by John Watrous (2018) for more information on quantum information theory. Much of content of this answer is similar to the answers to this related question; fedja remarked that the idea behind the answer to that previous question also applies to this question, so fedja could have answered at least most of this question before I did.
A function $\mathcal{E}:M_d(\mathbb{C})\rightarrow M_d(\mathbb{C})$ is said to be positive if $\mathcal{E}(P)$ is positive semidefinite whenever $P$ is positive semidefinite. If $W$ is a finite dimensional vector space, then let $L(W)$ be the set of all homomorphisms from $W$ to $W$. We say that $\mathcal{E}:M_d(\mathbb{C})\rightarrow M_d(\mathbb{C})$ is completely positive if whenever $V$ is a finite dimensional complex Hilbert space, the mapping $\mathcal{E}\otimes 1_V:L(\mathbb{C}^d\otimes V)\rightarrow L(\mathbb{C}^d\otimes V)$ is positive. We say that a linear mapping $\mathcal{E}:M_d(\mathbb{C})\rightarrow M_d(\mathbb{C})$ is trace preserving if $\text{Tr}(\mathcal{E}(A))=\text{Tr}(A)$ whenever $A\in M_d(\mathbb{C})$. We say that a linear mapping $\mathcal{E}:M_d(\mathbb{C})\rightarrow M_d(\mathbb{C})$ is a quantum channel if it is both completely positive and trace preserving.
If $X_1,\dots,X_r$ are complex matrices, then define a mapping $\Phi(X_1,\dots,X_r):M_{d}(\mathbb{C})\rightarrow M_{d}(\mathbb{C})$ by letting
$$\Phi(X_1,\dots,X_r)(X)=X_1XX_1^*+\dots+X_rXX_r^*.$$ The completely positive mappings from $M_d(\mathbb{C})$ to $M_d(\mathbb{C})$ are precisely the mappings of the form
$\Phi(X_1,\dots,X_r)$, and the quantum channels $\mathcal{E}:M_d(\mathbb{C})\rightarrow M_d(\mathbb{C})$ are precisely the mappings of the form $\Phi(X_1,\dots,X_r)$ where $X_1^*X_1+\dots+X_r^*X_r=1_d$.
If $\mathcal{E}$ is a quantum channel, then $\rho(\mathcal{E})=1$; quantum channels are a lot like stochastic matrices.
I claim that $\rho(X_1\dots X_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(X_1,\dots,X_r))$ whenever $X_1,\dots,X_r$ are $d\times d$-complex matrices.
Let $O_{d,r}$ be the set of all $(X_1,\dots,X_r)\in M_d(\mathbb{C})^r$ such that
$\Phi(X_1,\dots,X_r)$ is not nilpotent. Let $E_{d,r}$ be the set of all $(X_1,\dots,X_r)\in M_d(\mathbb{C})^r$ such that
$\Phi(\lambda BX_1B^{-1},\dots,\lambda BX_rB^{-1})$ is a quantum channel for some complex $\lambda\neq 0$ and invertible matrix $B$. Then by my answer to my other question, $E_{d,r}$ is a dense subset of $O_{d,r}$, and clearly $O_{d,r}$ is dense in $M_d(\mathbb{C})^r$.
Observe that $$\frac{\rho(Y_1\dots Y_r)^{2/r}}{\rho(\Phi(Y_1,\dots,Y_r))}=\frac{\rho(\lambda BY_1B^{-1}\dots \lambda BY_rB^{-1})^{2/r}}{\rho(\Phi(\lambda BY_1B^{-1},\dots,\lambda BY_rB^{-1}))}$$ whenever $\lambda\neq 0$, $B$ is invertible, and $\Phi(Y_1,\dots,Y_r)$ is not nilpotent. Therefore, if $\rho(Z_1\dots Z_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(Z_1,\dots,Z_r))$ whenever $\Phi(Z_1,\dots,Z_r)$ is a quantum channel, then $\rho(Y_1\dots Y_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(Y_1,\dots,Y_r))$ whenever $(Y_1,\dots,Y_r)\in E_{d,r},$
so because $E_{d,r}$ is dense in $M_{d}(\mathbb{C})^{r}$, we will be able to conclude that $\rho(X_1\dots X_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(X_1,\dots,X_r))$ whenever $X_1,\dots,X_r\in M_d(\mathbb{C})^r$.
If $\Phi(Z_1,\dots,Z_r)$ is a quantum channel, then $$d=\text{Tr}(1_d)=\text{Tr}(Z_1^*Z_1+\dots+Z_r^*Z_r)=\|Z_1\|^2_2+\dots+\|Z_r\|_2^2$$ where $\|\cdot\|_2$ denotes the Frobenius norm.
Therefore, by the arithmetic-geometric mean inequality, we have our main inequality, namely
$$\rho(Z_1\dots Z_r)^{2/r}\leq\|Z_1\dots Z_r\|_\infty^{2/r}\leq(\|Z_1\|_\infty^2\dots\|Z_r\|_\infty^2)^{1/r}\leq(\|Z_1\|_2^2\dots\|Z_r\|_2^2)^{1/r}\leq\frac{\|Z_{1}\|_2^2+\dots+\|Z_r\|_2^2}{r}
=\frac{d}{r}=\frac{d}{r}\rho(\Phi(Z_1,\dots,Z_r)).$$
Furthermore, if $\rho(Z_1\dots Z_r)^{2/r}=\frac{d}{r}\rho(\Phi(Z_1,\dots,Z_r))$,
then $\|Z_j\|_\infty^2=\|Z_j\|^2_2=d/r$ for all $j$. However, The only way to get $\|Z_{j}\|_\infty=\|Z_j\|_2$ is if $\text{Rank}(Z_j)=1$ since $\|W\|_{\infty}=\|W\|_2$ precisely when $\text{Rank}(W)=1$.
Therefore, for all $j$, there are column vectors $u_j,v_j$ such that $Z_j=u_jv_j^*$.
In this case, $$\|Z_j\|_2^2=\text{Tr}((u_jv_j^*)(u_jv_j^*)^*)=\text{Tr}(u_jv_j^*v_ju_j^*)=
\text{Tr}(u_j^*u_jv_j^*v_j)=\|u_j\|^2\cdot\|v_j\|^2.$$ Therefore, $\|Z_j\|_2=\|u_j\|\cdot\|v_j\|=\sqrt{d/r}$.
$$\rho(Z_1\dots Z_r)=|\text{Tr}(Z_1\dots Z_r)|=|\text{Tr}(u_1v_1^*\dots u_rv_r^*)|
=|\text{Tr}(v_1^*u_2\dots v_r^*u_1)|$$
$$=|\langle v_1,u_2\rangle|\dots|\langle v_r,u_1\rangle|.$$
Since $\rho(Z_1\dots Z_r)=\|Z_1\|_\infty\dots\|Z_r\|_\infty$, we have
$$|\langle v_1,u_2\rangle|\dots|\langle v_r,u_1\rangle|=\|u_1\|\cdot\|v_1\|\dots \|u_r\|\cdot\|v_r\|.$$
This means that there are constants $\gamma_1,\dots,\gamma_r$ where
$v_j=\overline{\gamma_j}\cdot u_{j+1}$. Therefore,
$Z_j=u_jv_j^*=u_j\gamma_j u_{j+1}^*=\gamma_j u_ju_{j+1}^*$.
Now, set $w_j=\|u_j\|\cdot v_{j}=\|u_j\|\cdot\overline{\gamma_j}u_{j+1}.$
Then $$1_d=\sum_{j=1}^{r}Z_j^*Z_j=\sum_{j=1}^{r}(u_jv_j^*)^*u_jv_j^*=\sum_{j=1}^{r}v_ju_j^*u_jv_j^*=\sum_{j=1}^{r}\|u_{j}\|^2\cdot v_jv_j^*=\sum_{j=1}^{r}w_jw_j^*.$$
Observe that $1_d=\sum_{j=1}^{r}w_jw_j^*$ precisely when $x=\sum_{j=1}^{r}w_{j}\cdot\langle x,w_j\rangle$. We say that a tuple of vectors $(h_1,\dots,h_r)$ is a normalized tight frame if $1_d=\sum_{j=1}^{r}h_jh_j^*$. We say that a normalized tight frame is equal-norm if $\|h_1\|=\dots=\|h_r\|$. Observe that $\|w_j\|^2=d/r$ for all $j$, so $(w_1,\dots,w_r)$ is an equal-norm tight frame.
One can also obtain tuples $(Z_1,\dots,Z_r)$ where $\Phi(Z_1,\dots,Z_r)$ is a quantum channel and where $\rho(Z_1,\dots,Z_r)^{2/r}=d/r$ from equal-norm tight frames. Suppose that $(w_1,\dots,w_r)\in(\mathbb{C}^d)^r$ is an equal-norm tight frame. Then we necessarily have $\|w_j\|^2_2=d/r$ for all $j$. Now, suppose that $|\alpha_j\beta_j|^2=r/d$ for $1\leq j\leq r$. Let $v_j=\alpha_jw_j,u_{j+1}=\beta_{j+1}w_j$, and as always, set $Z_j=u_jv_j^*$ for $1\leq j\leq r$. Then the reader can verify that $\Phi(Z_1,\dots,Z_r)$ is a quantum channel with $\rho(Z_1,\dots,Z_r)^{2/r}=\frac{d}{r}$ and every quantum channel $\Phi(Z_1,\dots,Z_r)$ with $\rho(Z_1,\dots,Z_r)^{2/r}=\frac{d}{r}$ is of this form.
The existence of equal-norm tight frames $(w_1,\dots,w_r)\in(\mathbb{R}^d)^r$ is already a known result. For example, the existence of such frames is Corollary 7.1 in An Introduction to Finite Tight Frames by Shayne F. D. Waldron.
By applying continuity, we can actually show that $\text{Rank}(X_j)=1$ even if we drop the assumption that $\Phi(X_1,\dots,X_r)$ is a quantum channel.
Let $f:O_{d,r}\rightarrow\mathbb{R}$ be the function where
$f(X_1,\dots,X_r)=\frac{\rho(X_1\dots X_r)^{2/r}}{\rho(\Phi(X_1,\dots,X_r))}.$
Let $g^-:M_d(\mathbb{C})\rightarrow\mathbb{R}$ be the function where
$g^-(X)=|\lambda_1\lambda_2|$ where $\lambda_1,\dots,\lambda_d$ are the eigenvalues of $X$ arranged in an order so that $|\lambda_1|\geq|\lambda_2|\geq\dots\geq|\lambda_d|$.
Define $g:O_{d,r}\rightarrow\mathbb{R}$ by letting $$g(X_1,\dots,X_r)=\frac{g^-(X_1)+\dots+g^-(X_r)}{r\cdot\rho(\Phi(X_1,\dots,X_r))}.$$
Suppose that $(X_1,\dots,X_r)\in O_{d,r}$ and $f(X_1,\dots,X_r)=\frac{d}{r}$. Then for each $\delta>0$, there is a neighborhood $U$ of $(X_1,\dots,X_r)$ where if
$(Y_1,\dots,Y_r)\in U$, then $f(Y_1,\dots,Y_r)>\frac{d}{r}-\delta$. Now, let $(Y_1,\dots,Y_r)\in U\cap E_{d,r}$. Then there is some $\lambda$ and invertible $B$ where if we set $Z_j=\lambda BY_jB^{-1}$ for all $j$, then $\Phi(Z_1,\dots,Z_r)$ is a quantum channel. Now, let $\lambda_{j,1},\dots,\lambda_{j,d}$ be the eigenvalues of $Z_j$ ordered in a way so that $|\lambda_{j,1}|\geq|\lambda_{j,2}|\geq\dots\geq|\lambda_{j,d}|$. Let $\sigma_{j,1}\geq\dots\geq\sigma_{j,d}$ denote the singular values of $Z_j$.
Then
$$\frac{d}{r}-\delta<f(Y_1,\dots,Y_r)=f(Z_1,\dots,Z_r)=\rho(Z_1\dots Z_r)^{2/r}
\leq(\sigma_{1,1}^2\dots\sigma_{r,1}^2)^{1/r}$$
$$\leq\frac{\sigma_{1,1}^2+\dots+\sigma_{r,1}^2}{r}
\leq\frac{(\sigma_{1,1}^2+\sigma_{1,2}^2)+\dots+(\sigma_{r,1}^2+\sigma_{r,2}^2)}{r}$$
$$\leq\frac{\|Z_1\|_2^2+\dots+\|Z_r\|_2^2}{r}=\frac{d}{r}.$$
We conclude that $\sigma_{j,1}^2\leq\frac{d}{r}$ and $\sigma_{j,2}^2\leq\delta$. Therefore, by Weyl's inequality, we have
$$|\lambda_{j,1}\lambda_{j,2}|\leq\sigma_{i,1}\sigma_{j,2}\leq\sqrt{\frac{d\delta}{r}}.$$
Therefore, we have
$$g(Y_1,\dots,Y_r)=g(Z_1,\dots,Z_r)\leq\sqrt{\frac{d\delta}{r}}.$$
We therefore conclude that $g(X_1,\dots,X_r)=0$ which means that
$\text{Rank}(X_j)\leq 1$ for $1\leq j\leq r$.
Best Answer
Yes. We have a necessary and sufficient characterization for when the Cauchy-Schwarz inequality becomes equality.
For this post, $U,V,W$ shall denote finite dimensional complex Hilbert spaces. Suppose that $A_1,\dots,A_n:U\rightarrow U,B_1,\dots,B_n:V\rightarrow V$ are linear operators. Then define an operator $\Gamma(A_1,\dots,A_n;B_1,\dots,B_n):L(V,U)\rightarrow L(V,U)$ by letting $$\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)(X)=A_1XB_1^*+\dots+A_nXB_n^*.$$
Observe that $\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)$ is similar to $A_1\otimes \overline{B_1}+\dots+A_n\otimes\overline{B_n}$ (Here, $\overline{C}=(C^{T})^{*}=(C^*)^T$, so $\overline{C}$ is the matrix obtained by replacing every entry in $C$ with its complex conjugate), so the spectral radius Cauchy-Schwarz inequality $$\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)) \leq\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}$$ hold in all cases. It is not too hard to prove the spectral radius Cauchy-Schwarz inequality by using the conventional Cauchy-Schwarz inequality and the characterization of $\rho(\Phi(A_1,\dots,A_n))^{1/2}$ given in the 1998 paper The $p$-norm joint spectral radius for even integers by Ding-Xuan Zhou.
Observe that if there is a $\lambda$ and an invertible $B$ with $B_j=\lambda BA_jB^{-1}$ for $1\leq j\leq n$, then $$\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)) =\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}.$$
I claim that when $(A_1,\dots,A_n)$ and $(B_1,\dots,B_n)$ have no invariant subspaces, this is the only way in which the spectral radius Cauchy-Schwarz inequality becomes equality. By decomposing $(A_1,\dots,A_n)$ and $(B_1,\dots,B_n)$ according to their invariant subspaces, we obtain necessary and sufficient conditions for when the Cauchy-Schwarz inequality is actually an equality.
Recall that a linear operator $\mathcal{E}:L(U)\rightarrow L(V)$ is called positive if $\mathcal{E}(P)$ is positive semidefinite whenever $P$ is positive semidefinite. A linear operator $\mathcal{E}:L(U)\rightarrow L(V)$ is called completely positive if $\mathcal{E}\otimes 1_W:L(U\otimes W)\rightarrow L(V\otimes W)$ is positive whenever $W$ is a finite dimensional Hilbert space. We say that an operator $\mathcal{E}:L(U)\rightarrow L(V)$ is trace preserving if $\text{Tr}(\mathcal{E}(X))=\text{Tr}(X)$ for all $X\in L(U)$. An operator $\mathcal{E}:L(U)\rightarrow L(V)$ is said to be a quantum channel if $\mathcal{E}$ is both trace preserving and completely positive.
The completely positive mappings from $L(V)$ to $L(V)$ are precisely the mappings of the form $\Phi(A_1,\dots,A_n)$. Observe that any linear mapping from $L(V,U)$ to $L(V,U)$ is of the form $\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)$. It is not too hard to show using the Frobenius inner product that the mapping $\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)$ is trace preserving if and only if $A_1^*B_1+\dots+A_n^*B_n=1_V$. In particular, the quantum channels are precisely the mappings of the form $\Phi(A_1,\dots,A_n)$ where $A_1^*A_1+\dots+A_n^*A_n=1_V$.
Observe that if $\mathcal{E}$ is a quantum channel, then $\rho(\mathcal{E})=1$.
Let $E_{U,n}$ be the collection of tuples $(A_1,\dots,A_n)\in L(U)^n$ such that there is some complex number $\lambda$ and invertible $B$ such that if $B_j=\lambda BA_jB^{-1}$ for $1\leq j\leq n$, then $\Phi(B_1,\dots,B_n)$ is a quantum channel. By this answer, $L(U)^n\setminus E_{U,n}$ is a quite small set whenever $n>1$. In particular, if $(A_1,\dots,A_n)$ has no-invariant subspace, then $(A_1,\dots,A_n)\in E_{U,n}$.
Theorem: Suppose that $(A_1,\dots,A_n)\in L(U)^n,(B_1,\dots,B_n)\in L(V)^n$ have no invariant subspace. Then $$\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n))=\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}$$ if and only if there is some invertible matrix $C$ along with some complex number $\eta\neq 0$ where $A_j=\eta CB_jC^{-1}$ for $1\leq j\leq n$.
Proof: Since $(A_1,\dots,A_n),(B_1,\dots,B_n)$ have no invariant subspace, there are non-zero complex numbers $\mu,\nu$ along with invertible matrices $A,B$ where if we set $R_j=\mu AA_jA^{-1},S_j=\nu BB_jB^{-1}$ for $1\leq j\leq n$, then $\mathcal{E}=\Phi(R_1,\dots,R_n),\mathcal{F}=\Phi(S_1,\dots,S_n)$ are quantum channels. Let $W$ be a complex inner product space with orthonormal basis $(e_1,\dots,e_n).$ Now, let $R,S\in L(V,V\otimes W)$ be the operators defined by letting $R=\sum_{j=1}^{n}R_j\otimes e_j,S=\sum_{j=1}^{n}S_j\otimes e_j$. Then $$\mathcal{E}(X)=\text{Tr}_{W}(RXR^*),\mathcal{F}(X)=\text{Tr}_{W}(SXS^*).$$
Define a mapping $\mathcal{G}$ by setting $$\mathcal{G}(X)=\text{Tr}_{W}(RXS^*)=\sum_{k=1}^nR_kXS_k^*.$$
Since $R_1^*R_1+\dots+R_n^*R_n=S_1^*S_1+\dots+S_n^*S_n=1_V$, the mappings $R,S$ are isometries.
Now, let $\lambda$ be an eigenvalue of $\mathcal{G}$. Then there is some eigenvector $X$ with $\mathcal{G}(X)=\lambda X$. Now perform a polar decomposition of $X$ to write $X=HP$ where $H$ is an isometry and $P$ is a positive semidefinite matrix. Therefore, we have $$\lambda HP=\lambda X=\mathcal{G}(X)=\text{Tr}_{W}(RXS^*)=\text{Tr}_W(RHPS^*).$$ Now, set $T=(H^*\otimes 1_W)RH$. Then $$\lambda P=\lambda H^*HP=H^*\text{Tr}_W(RHPS^*)=\text{Tr}_W((H^*\otimes 1_W)RHPS^*)=\text{Tr}_W(TPS^*).$$ Now let $P=\sum_{k=1}^{n}\sigma_ke_ke_k^*$. Then $\text{Tr}(\lambda P)=\lambda\cdot\sum_{k=1}^n\sigma_k$. Therefore, we have $$\text{Tr}(\lambda P)=\text{Tr}(\text{Tr}_W(TPS^*))=\text{Tr}(TPS^*)=\sum_{k=1}^{n}\sigma_k\text{Tr}(Te_ke_k^*S^*)=\sum_{k=1}^n\sigma_k\text{Tr}(Te_k(Se_k)^*)=\sum_{k=1}^n\sigma_k\langle Te_k,Se_k\rangle.$$
Therefore, since $$\lambda\cdot\sum_{k=1}^n\sigma_k=\sum_{k=1}^n\sigma_k\langle Te_k,Se_k\rangle,$$ we know that $|\lambda|\leq 1$. Furthermore, if $|\lambda|=1$, then we know that $Te_k=\lambda Se_k$ whenever $\sigma_k>0$. Therefore, we have $T|_{\text{Im}(P)}=\lambda\cdot S|_{\text{Im}(P)}$. In this case, we have $\lambda P=\text{Tr}_W(TPS^*)=\text{Tr}_W(\lambda SPS^*)$, so $P=\text{Tr}_W(SPS^*)$. Since $P=\text{Tr}_W(SPS^*)$ and since $(S_1,\dots,S_n)$ has no invariant subspace, we know that $P$ is positive semidefinite. Therefore, $(H^*\otimes 1_W)RH=T=\lambda S$.
Now, if $v\in V$, then $\|RH v|=\|v\|=\|\lambda Sv\|=\|(H^*\otimes 1_W)RHv\|$. Therefore, if $v\in V$, then $RHv\in\text{Im}(H)\otimes 1_W$. Therefore, since $\text{Im}(H)$ is a non-trivial invariance subspace of $U$, we know that $\text{Im}(H)=U$. Therefore, $H$ is a unitary operator.
Thus, $H^*R_jH=\lambda S_j$ for $1\leq j\leq n$. Thus, $H^*\mu AA_jA^{-1}H=\lambda \nu BB_jB^{-1}$. We conclude that $$A_j=\mu^{-1}\lambda\nu A^{-1}HBB_jB^{-1}H^{-1}A=\mu^{-1}\lambda\nu A^{-1}HBB_j(A^{-1}HB)^{-1}.$$
Q.E.D.
Theorem: Suppose that $A_1,\dots,A_n:U\rightarrow U,B_1,\dots,B_n:V\rightarrow V$ be linear operators. Then assign $U,V$ bases so that $$A_j=\begin{bmatrix} A_{j,1,1}&\dots&A_{j,1,u}\\ \vdots&\ddots&\vdots\\ A_{j,u,1}&\dots&A_{j,u,u} \end{bmatrix}$$ and $$B_j=\begin{bmatrix} B_{j,1,1}&\dots&B_{j,1,v}\\ \vdots&\ddots&\vdots\\ B_{j,v,1}&\dots&B_{j,v,v} \end{bmatrix}$$ and where for each $\alpha,\beta$ each of the $n$ matrices $A_{1,\alpha,\beta},\dots,A_{n,\alpha,\beta}$ have the same dimensions, for each $\alpha,\beta$, each of the $n$ matrices $B_{1,\alpha,\beta},\dots,B_{n,\alpha,\beta}$ have the same dimensions, and where $A_{j,\alpha,\beta}=0$ whenever $\alpha>\beta$, and where $B_{j,\alpha,\beta}=0$ whenever $\alpha>\beta$, and where for $1\leq\alpha\leq u$, the matrices $(A_{1,\alpha,\alpha},\dots,A_{n,\alpha,\alpha})$ have no non-trivial invariant subspace, and where if $1\leq\beta\leq v$, the matrices $(B_{1,\beta,\beta},\dots,B_{n,\beta,\beta})$ have non-trivial no invariant subspace either. Then $\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n))=\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}$ if and only if there are $\alpha,\beta$ with $1\leq\alpha\leq u,1\leq\beta\leq v$ and where
$\rho(\Phi(A_1,\dots,A_n))=\rho(\Phi(A_{1,\alpha,\alpha},\dots,A_{n,\alpha,\alpha}))$
$\rho(\Phi(B_1,\dots,B_n))=\rho(\Phi(A_{1,\beta,\beta},\dots,A_{n,\beta,\beta}))$, and
There is an invertible matrix $C$ and some complex number $\lambda\neq 0$ such that $A_{j,\alpha,\beta}=\lambda CB_{j,\alpha,\beta}C^{-1}$ for $1\leq j\leq n$.
A few observations:
In this answer, we actually have a couple of different proofs of the spectral radius Cauchy-Schwarz inequality. To prove the Cauchy-Schwarz inequality, we need to prove that $\rho(\Gamma(R_1,\dots,R_n;S_1,\dots,S_n))\leq 1$ whenever $\Phi(R_1,\dots,R_n),\Phi(S_1,\dots,S_n)$ are quantum channels, and the general case will follow from the fact that $E_{V,n}$ is dense in $L(V)^n$ whenever $n>1$. We have already shown that $\rho(\Gamma(R_1,\dots,R_n;S_1,\dots,S_n))\leq 1$, but there is another way of showing this using the induced trace norm.
If $\mathcal{H}:L(V)\rightarrow L(V)$, then define the induced trace norm of $\mathcal{H}$ to be $\|\mathcal{H}\|_1=\max\{\|\mathcal{H}(X)\|_1:\|X\|_1\leq 1\}.$ Recall that if $R,S$ are isometries and $A$ is a complex matrix, then $\|RAS^*\|_1=\|A\|_1$ whenever the matrix multiplication exists. Also, recall that $\|\text{Tr}_{V}(X)\|_1\leq\|X\|_1$ whenever this inequality makes sense.
We have $$\|\Gamma(R_1,\dots,R_n;S_1,\dots,S_n)(X)\|_1=\|\text{Tr}_{W}(RXS^*)\|_1\leq\|RXS^*\|_1=\|X\|_1.$$ Therefore, $\|\Gamma(R_1,\dots,R_n;S_1,\dots,S_n)\|_1\leq 1$, so $\rho(\Gamma(R_1,\dots,R_n;S_1,\dots,S_n))\leq 1$ since the induced trace norm is submultiplicative.
Empirical verification
I have empirically verified using computer calculations that the conclusions that we have made are reasonable. Define a fitness function $F:M_d(\mathbb{C})^n\times M_d(\mathbb{C})^n\rightarrow\mathbb{R}$ by letting $$F(A_1,\dots,A_n;B_1,\dots,B_n)=\frac{\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n))}{\rho(\Phi(A_1,\dots,A_n))^{1/2}\rho(\Phi(B_1,\dots,B_n))^{1/2}}.$$
One can maximize the value of $F(A_1,\dots,A_n;B_1,\dots,B_n)$ using gradient ascent to obtain examples of tuples $(A_1,\dots,A_n;B_1,\dots,B_n)$ with $F(A_1,\dots,A_n;B_1,\dots,B_n)\approx 1$, but in each of these examples, we always have a complex number $\lambda$ along with some invertible $B$ where $B_j\approx \lambda BA_jB^{-1}$ for $1\leq j\leq n$.