The previous answer contained a major error/misconception, and I apologize for the dealy in correcting it. The answer to the question is "yes" locally in the Zariski topology but "no" globally.
Comparison of Milnor K-cohomology and motivic cohomology via edge maps:
Motivic cohomology has Zariski descent and hence there is a
descent/hypercohomology/Brown-Gersten-Quillen/coniveau spectral sequence computing the motivic cohomology of a smooth scheme $X$. On the $E_1$-page, the Gersten resolution of the sheaves $\mathcal{H}^q(\mathbb{Z}(n))$ (Zariski sheafification of motivic cohomology) appears. The entry $E_1^{p,q}$ is $\bigoplus_{z\in X^{(p)}}{\rm H}^{q-p}(\kappa(z),\mathbb{Z}(n-p))$ and the differential $d_1:E_1^{p,q}\to E_1^{p+1,q}$ is given by the appropriate residue maps.
The $E_2$-spectral sequence has the form
$$
E^{p,q}_2={\rm H}^p(X,\mathcal{H}^q(\mathbb{Z}(n)))\Rightarrow {\rm H}^{p+q}(X,\mathbb{Z}(n))
$$
and the differentials go $d_s^{p,q}:E^{p,q}_s\to E^{p+s,q-s+1}_s$.
This spectral sequence can then be used to relate Milnor K-cohomology to motivic cohomology. The relevant thing to note is the identification
$$
\tau_{\geq s}\mathbb{Z}(s)_X\to\mathbf{K}^{\rm M}_{s,X}[-s]
$$
which can be found e.g. in Section 3.1, Corollary 3.2 of
- K. Rülling and S. Saito. Higher Chow groups with modulus and relative Milnor K-theory. arXiv:1504.02669v2.
This statement basically says that $\mathcal{H}^q(\mathbb{Z}(n))=0$ for $q>n$ and $\mathcal{H}^n(\mathbb{Z}(n))=\mathbf{K}^{\rm M}_n$, where $\mathcal{H}^q$ denotes the motivic cohomology sheaves. (It also says that the answer is yes locally in the Zariski topology.)
From this vanishing statement, we get edge maps ${\rm H}^p(X,\mathbb{Z}(n))\to {\rm H}^{p-n}(X,\mathcal{H}^n(\mathbb{Z}(n)))$ because in the line $q=n$ there are no incoming differentials and so $E^{p,n}_\infty$ is a subgroup of ${\rm H}^{p-n}(X,\mathcal{H}^n(\mathbb{Z}(n)))$. Since $E^{p,n}_\infty$ is the last subquotient of the filtration of ${\rm H}^p(X,\mathbb{Z}(n))$ this provides the above edge maps. These would be the natural comparison maps (and they are also mentioned in Thi's answer.)
Comparison isomorphisms for $r=s$ and $r=s-1$:
To get the Bloch formula and the identification of ${\rm H}^{2n-1}(X,\mathbb{Z}(n))\cong {\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)$, we need to show that there can be no outgoing differentials either. In the $E_2$-page, the relevant differentials for Bloch's formula are $d_s^{p,q}:{\rm H}^n(X,\mathcal{H}^n(\mathbb{Z}(n)))\to {\rm H}^{n+s}(X,\mathcal{H}^{n-s+1}(\mathbb{Z}(n)))$. The latter group is trivial, since on the $E_1$-page, the coefficients would be $\mathcal{H}^{1-2s}(\mathbb{Z}(-s))$ and the sheaves $\mathbb{Z}(-s)$ are trivial for $s>0$. The other relevant differentials are $d_s^{p,q}:{\rm H}^{n-1}(X,\mathcal{H}^n(\mathbb{Z}(n)))\to {\rm H}^{n+s-1}(X,\mathcal{H}^{n-s+1}(\mathbb{Z}(n)))$ with the coefficients in the $E_1$-page given by ${\rm H}^{2-2s}(\mathbb{Z}(1-s))$. For $s>1$, this is again trivial. In particular, all the outgoing differentials are trivial, so that $E^{n,n}_\infty={\rm H}^n(X,\mathbf{K}^{\rm M}_n)$ and $E^{n-1,n}_\infty={\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)$. This also implies that the only $E^{n+s,n-s}_\infty$ contributing to ${\rm H}^{2n}(X,\mathbb{Z}(n)))$ is ${\rm H}^n(X,\mathbf{K}^{\rm M}_n)$, yielding Bloch's formula. To get ${\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)\cong {\rm H}^{2n-1}(X,\mathbb{Z}(n))$ we need to show that ${\rm H}^{n+s-1}(X,\mathcal{H}^{n-s}(\mathbb{Z}(n)))=0$. The relevant coefficients in the $E_1$-page are $\mathcal{H}^{1-2s}(\mathbb{Z}(1-s))$. For $s=1$, we get $\mathcal{H}^{-1}(\mathbb{Z}(0))$ which is trivial, all the other terms are again trivial because $\mathbb{Z}(1-s)=0$ for $s>1$.
Failure of global comparison:
The natural comparison isomorphism fails to be an isomorphism in general. The simplest counterexample (pointed out by Jens Hornbostel) is actually $\mathbb{P}^1$. In this case, we have
$$
{\rm H}^r(\mathbb{P}^1,\mathbf{K}^{\rm M}_s)\cong \left\{\begin{array}{ll} {\rm K}^{\rm M}_s(F) & r=0\\
{\rm K}^{\rm M}_{s-1}(F) & r=1\end{array}\right.
$$
where $F$ denotes the base field. On the other hand, the projective bundle formula for motivic cohomology implies
$$
{\rm H}^{r+s}(\mathbb{P}^1,\mathbb{Z}(s))\cong {\rm H}^{r+s}(F,\mathbb{Z}(s))\oplus {\rm H}^{r+s-2}(F,\mathbb{Z}(s-1)).
$$
For $r=0$, we have ${\rm H}^0(\mathbb{P}^1,\mathbf{K}^{\rm M}_s)\cong {\rm K}^{\rm M}_s(F)$ and ${\rm H}^s(\mathbb{P}^1,\mathbb{Z}(s))\cong {\rm K}^{\rm M}_s(F)\oplus {\rm H}^{s-2}(F,\mathbb{Z}(s-1))$, so Milnor K-cohomology and motivic cohomology differ whenever ${\rm H}^{s-2}(F,\mathbb{Z}(s-1))\neq 0$. A particular instance where that happens is $s=3$ in which case ${\rm H}^1(F,\mathbb{Z}(2))\cong {\rm K}^{\rm ind}_3(F)$.
In the formulation of the descent spectral sequence, the spectral sequence is contained in the two columns for $p=0,1$ and degenerates at the $E_2$-term. The failure of the global comparison follows since $\mathcal{H}^2(\mathbb{Z}(3))$ has nontrivial first cohomology over $\mathbb{P}^1$, given by ${\rm H}^1(F,\mathbb{Z}(2))$.
Best Answer
The answer is no.
So far, you have checked only very special cases, such as weight $n$ and degree $2n$, for $n = 0,1$.
Consider the analytic hypercohomology of your $\mathbf{Z}(n)_{\mathcal{M}}$ in degree $2n$, denoted $H^{2n}_{\rm an}(X,\mathbf{Z}(n))$. The same (elementary) argument as in several references (eg. Bloch's "Algebraic Cycles and Higher $K$-Theory", 1986) shows
$$H^{2n}_{\rm an}(X,\mathbf{Z}(n)) \simeq\text{CH}^n(X^{\rm an})$$ the Chow group of analytic cycles on $X^{\rm an}$. By GAGA, there is a canonical isomorphism of abelian groups $\text{CH}^n(X)\simeq\text{CH}^n(X^{\rm an})$.
Upon running the same construction of the cycle map to Deligne cohomology as in Kerr-Lewis, as you suggest, you must end up with a cycle map
$$\text{CH}^n(X^{\rm an})\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$$ that, under the above isomorphism from GAGA, should better agree with the usual cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$.
Rem. If your conjecture were true, then, in particular, the cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$ should be surjective, this latter group being surjecting onto the finitely generated abelian group of Hodge cycles $\text{Hdg}^{n,n}(X^{\rm an})$, the composition being the cycle map to Betti cohomology. As a consequence, the Hodge conjecture is implied by your contention.
There are two problems. I am not aware of any expectation along the lines of surjectivity of the cycle map onto Deligne(-Beilinson) cohomology. Rather, the only claim being usually made is surjectivity onto the abelian group of Hodge classes.
On the other hand, if your contention were correct, then the cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$ should be injective too, which rarely happens, as this puts strong constraints on the Hodge numbers of $X^{\rm an}$.
A paper of Esnault and Levine proves that if such cycle map is injective, then it is also surjective, and implies strong conditions on the shape of the Hodge diamond of $X^{\rm an}$. There are many examples of an $X$ that does not meet such conditions.