$\newcommand{\F}{\mathcal{F}}$The answer is yes.
Indeed, by time rescaling, without loss of generality (wlog) $r=1$.
Take any random variable (r.v.) $Z$ with $EZ=0$, and let $Z'$ be an independent copy of $Z$. Then, by Jensen's inequality, for any real $z$ we have $E|Z|\le E|Z-Z'|=E|(Z-z)-(Z'-z)|\le2E|Z-z|$, so that
\begin{equation*}
E|Z|\le2E|Z-z|. \tag{1}
\end{equation*}
Letting $E_t:=E(\cdot|\F_t)$ and using (1) with $Z=X_t-X_{t-1}$ and real $t\ge1$, we get
\begin{equation*}
\begin{aligned}
E_{t-1}|X_t-X_{t-1}|&\le2E_{t-1}|(X_t-X_{t-1})-(Y_t-X_{t-1})| \\
&=2E_{t-1}|X_t-Y_t|,
\end{aligned}
\end{equation*}
since $Y_t-X_{t-1}$ is $\F_{t-1}$-measurable.
So, $E|X_t-X_{t-1}|\le2|X_t-Y_t|$. So, for any real $u\ge0$ and any natural $T$,
\begin{equation*}
S_u:=\sum_{n=T}^\infty E|X_{n+u}-X_{n-1+u}|\le2\sum_{n=T}^\infty E|X_{n+u}-Y_{n+u}|
\end{equation*}
and hence
\begin{equation*}
\begin{aligned}
\int_0^1 du\, S_u &\le2\sum_{n=T}^\infty \int_0^1 du\, E|X_{n+u}-Y_{n+u}| \\
& =2\int_T^\infty dt\, E|X_t-Y_t|<\infty
\end{aligned}
\end{equation*}
if $T$ is large enough, by your displayed condition. So, $S_u<\infty$ for some $u\in[0,1]$. By time shift, wlog $u=1$. So,
\begin{equation*}
\sum_{n=T}^\infty E|X_{n+1}-X_n|=S_1<\infty. \tag{2}
\end{equation*}
So, $(X_n)$ converges in $L^1$ and hence $(E|X_n|)$ is bounded. So, by Doob's martingale convergence theorem,
\begin{equation*}
X_n\to Y \tag{3}
\end{equation*}
as $n\to\infty$ almost surely (a.s.) for some (real-valued) r.v. $Y$.
Next, by Doob's martingale inequality,
\begin{equation*}
P(\max_{t\in[n,n+1]}|X_t-X_n|>h)\le\frac{E|X_{n+1}-X_n|}h
\end{equation*}
for any real $h>0$.
So, by the Borel–Cantelli lemma and (2), for each real $h>0$ a.s. there will be only finitely many natural $n$ such that $\max_{t\in[n,n+1]}|X_t-X_n|>h$. That is, $\max_{t\in[n,n+1]}|X_t-X_n|\to0$ a.s. as $n\to\infty$.
Thus, by (3), $X_t\to Y$ a.s. as $t\to\infty$, as claimed.
If $\xi\in \mathbb D^{1,2}$ is in the Sobolev-Watanabe space then we can apply Clark-Ocone formula to get that
$$\xi=E[\xi]+\int_0^T E[D_s\xi|\mathcal F_s]dW_s$$
where $D_s$ is the Malliavin derivative. For $s\in [0,T]$ we may write $\xi=\xi 1_{\{\tau > s\}}+\xi 1_{\{\tau \leq s\}}$. Then
\begin{align*}
\xi&=E[\xi]+\int_0^T E[D_s(\xi 1_{\{\tau > s\}}+\xi 1_{\{\tau \leq s\}})|\mathcal F_s]dW_s\\
&=E[\xi]+\int_0^T E[D_s(\xi 1_{\{\tau > s\}})|\mathcal F_s]dW_s+\int_0^T E[D_s(\xi 1_{\{\tau \leq s\}})|\mathcal F_s]dW_s
\end{align*}
$\xi 1_{\{\tau \leq s\}}$ is $\mathcal F_s$-measurable so $D_s (\xi 1_{\{\tau \leq s\}})=0$. Also for $s>\tau$ we have that $\xi 1_{\{\tau > s\}}=0$ so $D_s (\xi 1_{\{\tau > s\}})=0$ and for $s<\tau$ we have that $\xi 1_{\{\tau > s\}}=\xi$. So
$$\xi=E[\xi]+\int_0^\tau E[D_s\xi|\mathcal F_s]dW_s.$$
Best Answer
I managed to work it out - the answer is yes.
By breaking $f$ into positive and negative parts, it suffices to prove the claim for positive $f$. By replacing $f$ with $\max(f, N)$ for $N > 0$ and applying the dominated convergence theorem as $N \to \infty$, we can further restrict to proving the claim for bounded, positive $f$.
Using the definition of the Riemann-Stiltjes integral, with $\mathcal P = \{a_i\}_{i = 0}^\infty $ denoting a generic partition of $\mathbb R_+$, we write
$$\mathbb E \left [\int_0^\infty f(X_t) \, dF_t \right ] = \mathbb E \left [\lim_{|\mathcal P| \to 0} \sum_{i = 0}^\infty f(X_{a_i}) [ \mathbb P(\tau \leq a_{i+1} | \mathcal F_{a_{i+1}}) - \mathbb P(\tau \leq a_i | F_{a_i})] \right ].$$
Since $f$ is bounded, we may swap the almost sure limit as $|\mathcal P| \to 0$ and the expectation, and then the sum and expectation and get
$$\lim_{|\mathcal P| \to 0} \sum_{i = 0}^\infty\mathbb E [ f(X_{a_i}) (\mathbb P(\tau \leq a_{i+1} | \mathcal F_{a_{i+1}}) - \mathbb P(\tau \leq a_i | \mathcal F_{a_i}))] $$
Now we perform an intermediate conditioning - conditioning the $i$'th summand on $\mathcal F_{a_i}$, we obtain
$$\lim_{|\mathcal P| \to 0} \sum_{i = 0}^\infty \mathbb E \left [\mathbb E \left [f(X_{a_i}) \mathbb 1_{\{a_i < \tau \leq a_{i+1}\}} \big | \mathcal F_{a_i} \right ] \right ]$$
$$=\lim_{|\mathcal P| \to 0} \sum_{i = 0}^\infty \mathbb E\left [ f(X_{a_i}) \mathbb 1_{\{a_i < \tau \leq a_{i+1}\}} \right ]$$
$$=\lim_{|\mathcal P| \to 0} \mathbb E\left [ \sum_{i = 0}^\infty f(X_{a_i}) \mathbb 1_{\{a_i < \tau \leq a_{i+1}\}} \right ]$$ so that
$$\mathbb E \left [\int_0^\infty f(X_t) \, dF_t \right ] - \mathbb E[f(X_\tau)]$$
$$ = \lim_{|\mathcal P| \to 0}\mathbb E\left [ \sum_{i = 0}^\infty f(X_{a_i}) \mathbb 1_{\{a_i < \tau \leq a_{i+1}\}} \right ] - \mathbb E[f(X_\tau)]$$
$$= \lim_{|\mathcal P| \to 0} \mathbb E\left [ \sum_{i = 0}^\infty (f(X_{a_i}) - f(X_\tau)) \mathbb 1_{\{a_i < \tau \leq a_{i+1}\}} \right ]$$
Now using continuity of $f$ and $X$, we deduce that the term inside the expectation converges to $0$ in probability, hence in $L^1$ by boundedness of $f$. This proves the claim.