THE MAIN RESULT
Theorem Employing the notation and terminology introduced in the original post, we have: $\Phi$ is $\mathcal{B}(\mathbf{C}_{(0)})\otimes\mathcal{B}[0,\infty)\otimes\mathcal{B}(\mathbf{C}_{(0)})/\mathcal{B}(\mathbf{C}_{(0)})$ measurable.
Proof (For a listing of pertinent results, notations and definitions consult the next section, "Auxiliary facts and definitions". References cited below - e.g. [K], [M] - are listed in full at the bottom of this post in the section "Works cited".)
I will prove only the case $d = 1$.
As suggested in [S] p. 72, it suffices to show that $\Phi$ is continuous w.r.t. the product topology $\mathcal{T}$ on $\mathbf{C}_{(0)} \times [0,\infty) \times \mathbf{C}_{(0)}$ where $\mathbf{C}_{(0)}$ is equipped with the $\mathcal{O}_\rho$ topology, and $[0,\infty)$ is equipped with the standard, Euclidean topology, since if $\Phi$ is $\mathcal{T}/\mathcal{O}_\rho$-continuous, then by lemma 16 ("continuity and measurability"), $\Phi$ is $\sigma(\mathcal{T})/\sigma(\mathcal{O}_\rho)$-measurable, but by lemma 15 ("product and Borel $\sigma$-fields) and in light of proposition 8 ("$\mathbf{C}_{(0)}$ is a Polish space") we have $\sigma(\mathcal{T}) = \sigma(\mathcal{O}_\rho)\otimes\mathcal{B}[0,\infty)\otimes\sigma(\mathcal{O}_\rho)$ and by definition 7, $\sigma(\mathcal{O}_p) = \mathcal{B}(\mathbf{C}_{(0)})$.
Now, to prove that $\Phi$ is $\mathcal{T}/\mathcal{O}_\rho$-continuous, let $(f^*, t^*, g^*) \in \mathbf{C}_{(0)} \times [0,\infty) \times \mathbf{C}_{(0)}$ and let $\varepsilon \in (0,\infty)$. We shall define some $\delta \in (0,\infty)$, such that for all $(f, t, g) \in \mathbf{C}_{(0)} \times [0,\infty) \times \mathbf{C}_{(0)}$ with $\rho(f, f^*), |t - t^*|, \rho(g, g^*) < \delta$, we will have
$$
\rho(\Phi(f,t,g), \Phi(f^*, t^*, g^*)) < \varepsilon
$$
Choose $k, n \in \mathbb{N}_1$ such that
a) $t^* + 1/8 \leq k$,
b) $(k + 1)2^{-k} \leq \varepsilon$
c) $k \leq n$
d) Whenever $s, s' \in [0, k]$ are such that $|s - s'| < 2^{- (n + 2)}$, $|g^*(s) - g^*(s')| < 2^{- (k + 2)}$
And define $\delta := 2^{- (n + 2)}$.
Let $(f, t, g) \in \mathbf{C}_{(0)} \times [0,\infty) \times \mathbf{C}_{(0)}$ be such that $\rho(f, f^*), |t - t^*|, \rho(g, g^*) < \delta$, and set
$$
\begin{aligned}
\phi & := \Phi(f,t,g) \\
\phi^* & := \Phi(f^*, t^*, g^*) \\
m & := \min(t, t^*) \\
M & := \max(t, t^*)
\end{aligned}
$$
Note that $M \leq k$.
We have
$$
\sup_{0 \leq s \leq k}|\phi(s) - \phi^*(s)| = \\
\max(\sup_{0 \leq s \leq m}|\phi(s) - \phi^*(s)|, \sup_{m \leq s \leq M}|\phi(s) - \phi^*(s)|, \sup_{M \leq s \leq k)} |\phi(s) - \phi^*(s)|)
$$
Now, by lemma 6a ("Equivalence of $\rho$ and $\sup$"),
$$
\sup_{0 \leq s \leq m}|\phi(s) - \phi^*(s)| = \sup_{0 \leq s \leq m}|f(s) - f^*(s)| < 2^{- (n + 2)} < 2^{-k}
$$
Secondly, we have (by lemma 6a and by the assumptions on $k$ and on $n$),
$$
\sup_{m \leq s \leq M} |\phi(s) - \phi^*(s)| \leq \sup_{m \leq s \leq M} |f(s) - f^*(s)| + \sup_{0 \leq s \leq 2^{- (n + 2)}} |g^*(s)| + \sup_{0 \leq s \leq 2^{- (n + 2)}} |g(s) - g^*(s)| \\
< \frac{2^{-n}}{4} + \frac{2^{-k}}{4} + \frac{2^{-n}}{4} < 2^{-k}
$$
Thirdly, we have (again, by lemma 6a and by the assumptions on $k$ and on $n$),
$$
\sup_{M \leq s \leq k} |\phi(s) - \phi^*(s)| \leq \sup_{M \leq s \leq k} |f(s) - f^*(s)| + \sup_{0 \leq s \leq k - M} |g(s) - g^*(s)| \\
+ \sup_{0 \leq s \leq k - m}|g^*(s + 2^{- (n + 2)}) - g^*(s)| < \frac{2^{-n}}{4} + \frac{2^{-n}}{4} + \frac{2^{-k}}{4} < 2^{-k}
$$
In conclusion,
$$
\sup_{0 \leq s \leq k} |\phi(s) - \phi^*(s)| < 2^{-k}
$$
Hence, by lemma 6b, $\rho(\phi, \phi^*) < (k + 1)2^{-k} \leq \varepsilon$.
Q.E.D.
AUXILIARY FACTS AND DEFINITIONS
Definition 1 ($\pi_t$, the canonical projection onto the $t$th coordinate). ([S], section 4.1, p. 40) Setting $I := [0,\infty)$, denote by $\pi_t : (\mathbb{R}^d)\rightarrow \mathbb{R}^d$, $\omega \mapsto \omega(t)$ the canonical projection onto the $t$th coordinate.
Definition 2 (The product $\sigma$-algebra $\mathcal{B}^I(\mathbb{R}^d)$). ([S], section 4.1, p. 40) The natural $\sigma$-algebra on the infinite product $(\mathbb{R}^d)^I$ is the product $\sigma$-algebra
$$
\mathcal{B}^I(\mathbb{R}^d) := \sigma \{\pi_t^{-1}(B) : B \in \mathcal{B}(\mathbb{R}^d), t \in I\} = \sigma \{\pi_t : t \in I\}
$$
Definition & Proposition 3 (The trace $\sigma$-algebra). (a: [S2] Example 3.3 (vi), p. 16; b: [H] Theorem E, p. 25) Let $\Omega$ be some non-empty set, let $\mathcal{F}$ be a $\sigma$-algebra over $\Omega$ and let $S \subseteq \Omega$.
a) $S \cap \mathcal{F} := \{S \cap T : T \in \mathcal{F}\}$ (called the trace $\sigma$-algebra of $S$ w.r.t. $\mathcal{F}$) is a $\sigma$-algebra over $S$.
b) If $\mathcal{E} \subseteq \mathcal{F}$ is a generator of $\mathcal{F}$ (i.e. if $\mathcal{F} = \sigma(\mathcal{E})$), then $S \cap \mathcal{F} = \sigma(S \cap \mathcal{E})$.
Corollary 4 ($\mathbf{C}_{(0)} \cap \mathcal{B}^I(\mathbb{R}^d) = \sigma(\pi_t\mid_{\mathbb{C}_{(0)}} : t \in I)$). ([S] section 4.1, p. 41)
$$
\mathbf{C}_{(0)} \cap \mathcal{B}^I(\mathbb{R}^d) = \sigma(\pi_t\mid_{\mathbb{C}_{(0)}} : t \in I)
$$
Definition & Proposition 5 ($\rho$, the locally uniform convergence metric). (Based on, but not identical to, [S] section 4.1, p. 41) The function $\rho: \mathbf{C}_{(0)} \rightarrow [0,\infty)$,
$$
\rho(w,v) := \sum_{n = 1}^\infty \left(2^{-n} \wedge \sup_{0 \leq t \leq n}|w(t) - v(t)|\right)
$$
is a metric on $\mathbf{C}_{(0)}$.
Lemma 6 (Equivalence of $\rho$ and $\sup$). Let $n \in \mathbb{N}_1$ and let $w,v \in \mathbf{C}_{(0)}$.
a. If $\rho(w,v) < 2^{-n}$, $\sup_{0 \leq t \leq n}|w(t) - v(t)| < 2^{-n}$.
b. If $\sup_{0 \leq t \leq n}|w(t) - v(t)| < 2^{-n}$, $\rho(w,v) < (n + 1) 2^{-n}$.
Proof of lemma 6
a. Assume that $\rho(w,v) < 2^{-n}$. Suppose, by way of contradiction, that $\sup_{0 \leq t \leq n}|w(t) - v(t)| \geq 2^{-n}$. Then
$$
\rho(w,v) \geq 2^{-n} \wedge \sup_{0 \leq t \leq n} |w(t) - v(t)| = 2^{-n}
$$
contrary to the assumption.
b. Assume that $\sup_{0 \leq t \leq n}|w(t) - v(t)| < 2^{-n}$. Then
$$
\rho(w, v) \leq \sum_{m = 1}^n 2^{-n} + \sum_{m = n+1}^\infty 2^{-m} = n2^{-n} + 2^{-n}
$$
Q.E.D.
Definition 7 ($\mathcal{O}_\rho$, $\mathcal{B}(\mathbf{C}_{(0)})$). ([S] section 4.1, p. 41) Denote by $\mathcal{O}_\rho$ the topology induced by $\rho$ and consider the Borel $\sigma$-algebra $\mathcal{B}(\mathbf{C}_{(0)}) := \sigma(\mathcal{O}_\rho)$ on $\mathbf{C}_{(0)}$.
Proposition 8 ($\mathbf{C}_{(0)}$ is a Polish space). ([S] section 4.1, p. 41) Equipped with the metric $\rho$ defined above, $\mathbf{C}_{(0)}$ becomes a Polish space, i.e. a complete, separable metric space.
Lemma 9 ($\mathbf{C}_{(0)}\cap\mathcal{B}^I(\mathbb{R}^d) = \mathcal{B}(\mathbf{C}_{(0)})$). ([S] Lemma 4.1, p. 41) We have
$$
\mathbf{C}_{(0)}\cap\mathcal{B}^I(\mathbb{R}^d) = \mathcal{B}(\mathbf{C}_{(0)})
$$
Proposition 10 (A Brownian motion $\omega \mapsto (t \mapsto B(\omega, t))$ is $\mathcal{F}/\mathbf{C}_{(0)}$-measurable). ([S] top paragraph on p. 42) A standard, $d$-dimensional Brownian motion $(B(t))_{t \geq 0}$ over the probability space $(\Omega, \mathcal{F}, P)$, when considered as a function $\in \Omega \mapsto \mathbf{C}_{(0)}$, is $\mathcal{F}/\mathcal{B}(\mathbf{C}_{(0)})$-measurable.
Definition 11 (Box topology). ([M] p. 114) Let $\{X_\alpha\_{\alpha \in J}$ be an indexed family of topological spaces. Let us take as a bases for a topology on the product space $\prod_{\alpha \in J}X_\alpha$ the collection of all sets of the form $\prod_{\alpha \in J}U_\alpha$ where $U_\alpha$ is open in $X_\alpha$, for each $\alpha \in J$. The topology generated by this basis is called the box topology.
Definition 12 (Product topology, product space). ([M] p. 114) Let $\mathcal{S}_\beta$ denote the collection
$$
\mathcal{S}_\beta := \{\pi_\beta^{-1}(U_\beta) \mid U_\beta \textrm{ open in } X_\beta \}
$$
and let $\mathcal{S}$ denote the union of these collections, $\mathcal{S} := \bigcup_{\beta \in J}\mathcal{S}_\beta$. The topology generated by the sub basis $\mathcal{S}$ is called the product topology. In this topology, $\prod_{\alpha \in J} X_\alpha$ is called a product space.
Theorem 13 (Comparison of the box and product topologies). ([M] Theorem 19.1, p. 115) The box topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$. The product topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$ and $U_\alpha$ equals $X_\alpha$ except for finitely many values of $\alpha$.
Corollary 14 (Equality of the box and product topologies for finite products). ([M] p. 115) For finite products $\prod_{\alpha = 1}^n X_\alpha$, the two topologies are precisely the same.
Lemma 15 (Product and Borel $\sigma$-fields). ([K] Lemma 1.2, p. 3) If $S_1, S_2, \dots$ are separable metric spaces with induced topologies $\mathcal{T}_1, \mathcal{T}_2, \dots$, respectively, and if $\mathcal{T}$ is the product topology on $S_1 \times S_2 \times \cdots$, then
$$
\sigma(\mathcal{T}) = \sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2) \otimes \cdots
$$
Lemma 16 (Continuity and measurability). ([K] Lemma 1.5, p. 4) Let $f$ be a continuous mapping between two topological spaces $S$ and $T$ with Borel $\sigma$-fields $\mathcal{S}$ and $\mathcal{T}$. Then $f$ is $\mathcal{S}/\mathcal{T}$-measurable.
WORKS CITED
[H] Halmos, Paul R. Measure Theory. Springer-Verlag, 1974
[K] Kallenberg, Olav. Foundations of Modern Probability. 2nd edition. Springer, 2002
[M] Munkres, James R. Topology. 2nd edition. Prentice Hall, 2000
[S] Schilling, René L. and Partzsch, Lothar. Brownian motion - an introduction to stochastic processes. De Gruyter, 2012
[S2] Schilling, René L. Measures, integrals and martingales. Cambridge University Press, 2011
Best Answer
Following up on my comment, we will show that given $A=[B_T \in I]$ for any interval $I$, we have $A \cap [T\le t] \in \mathcal{F}^+(t)$ for all $t$. To take this to a Borel set of $\mathbb{R}$ rather than an interval, one can use standard measurable theoretic arguments.
First, as $T$ is a stopping time, we have $[T\le t] \in \mathcal{F}(t) \subset \mathcal{F}^+(t) $, this is the definition of stopping time.
Now, as $T\le t$, we have that $(B_T)(\omega) = B^t_{\cdot}(\omega) \circ T(\omega)$ where $B^t_s(\omega):= B_{t \wedge s}(\omega)$.
As $t \mapsto B_t$ is continuous (and therefore measurable) and $\{B^t_s: s \ge 0\}$ is measurable in $\mathcal{F}^+(t)$.
Using that given two functions $f: \mathbb{R}\times \Omega \mapsto \mathbb{R}$ and $g: \Omega \mapsto \mathbb{R}$, the composition $F(\omega)=f(g(\omega),\omega)$ is also measurable, we complete the answer. One can prove this last statement by taking $g$ to be a simple function first, then taking an approximation via simple function for general $g$.