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
Here is my own solution. Reading The answer from TheBridge, and not being able to completely follow the line of reasoning, I made a new attempt and - barring I'm in error - was able to solve it myself. (I think this just is the same argument TheBridge is making.)
In particular it does not seem that we need the Strong Markov Property.
First let us determine the distribution of $W_t := \Phi(B(t \wedge \tau), \tau, B(\cdot + \tau) - B(\tau))$: Given any $s >0$, consider the partition of $\Omega$ by the sets $$ A_1=\{\tau \le s \} $$ and $$ A_2 = \{ s < \tau \}. $$ For some arbitrary set $B \in \mathcal B (\mathbb R)$, Let us consider $$ \left\{ W_s \in B \right\} = \bigcup_{i=1}^2 \left \{ W_s \in B \cap A_i \right \}. $$ For $\omega \in A_1$ we have that $\tau (\omega) \le s$ and hence from the definition of $\mathrm \Phi$, \begin{align*} W_s(\omega) &= \underbrace{B (\tau(\omega)\wedge\tau(\omega), \omega )}_{f(\tau(\omega))} + \underbrace{B((s - \tau(\omega)) + \tau(\omega), \omega) - B(\tau(\omega), \omega)}_{g(s - \tau(\omega))} \\ &= B(s, \omega). \end{align*} Thus $$ \left \{ W_s \in B \cap A_1 \right \} = \left \{ B_s \in B \cap A_1 \right \}. $$ For $\omega \in A_2$ we have that $s < \tau(\omega)$ and hence from the definition of $\mathrm \Phi$, \begin{align*} W_s(\omega) &= B (s \wedge \tau(\omega), \omega ) \\ &= B (s , \omega ). \end{align*} Thus $$ \left \{ W_s \in B \cap A_2 \right \} = \left \{ B_s \in B \cap A_2 \right \}, $$ and we conclude \begin{align*} \left\{ W_s \in B \right\} &= \bigcup_{i=1}^2 \left \{ W_s \in B \cap A_i \right \} \\ &= \bigcup_{i=1}^2 \left \{ B_s \in B \cap A_i \right \} \\ &= \left \{ B_s \in B \right \}, \end{align*} and hence $$ \mathrm P \left\{ W_s \in B \right\} = \mathrm P \left \{ B_s \in B \right \} $$ Now, consider $s_1 < s_3$ and the increment $W_{s_2} - W_{s_1}$. By partitioning $\Omega $ into $$ \{ s_2 < \tau \} \bigcup \{ s_1 < \tau \le s_2 \} \bigcup \{ \tau \le s_1 \}, $$ it follows from an analogous argument that $$ W_{s_2} - W_{s_1} \sim B_{s_2} - B_{s_1}. $$
Further, the same kind of argument, and using that $B$ has independent increment, also works to show that for any $0 \le s_1 < s_2 < s_3 < s_4$, and any $A, B \in \mathcal{B}(\mathbb R)$ \begin{align*} \mathrm P \left \{ W_{s_2} - W_{s_1} \in A, W_{s_4} - W_{s_3} \in B \right \} &= \mathrm P \left \{ B_{s_2} - B_{s_1} \in A, B_{s_4} - B_{s_3} \in B \right \} \\ &= \mathrm P \left \{ B_{s_2} - B_{s_1} \in A\right\} \mathrm P\left \{ B_{s_4} - B_{s_3} \in B \right \} \\ &= \mathrm P \left \{ W_{s_2} - W_{s_1} \in A\right\} \mathrm P\left \{ W_{s_4} - W_{s_3} \in B \right \}. \end{align*} Hence $W$ has independent increments.