I think the following example says no.

Consider the state space $\{0,1\}$. Let $U$ be a uniform random variable on $[0,1]$ and let $X_t = 1$ if $t=U$ and $X_t = 0$ otherwise. Note that $X_t$ is a.s. not cadlag, but for each $t$ we have $X_t = 0$ a.s., so $X_t$ is a modification of the cadlag process $\hat{X}_t$ that just sits at 0. Taking $\mathcal{G}_t = \sigma(X_s : s \le t)$, unless I am mistaken, $X_t$ meets the Revuz/Yor definition of a Markov process (under its natural filtration). We may take the transition semigroup to simply be the identity $P_t f = f$, which is Feller.

But if we let $\tau = \inf\{t : X_t \ne 1\}$ and let $f(0)=0$, $f(1)=1$, then $E[f(X_\tau)] = 1$ while since $\hat{\tau} = +\infty$ we might consider $E[f(\hat{X}_{\hat{\tau}})] = 0$.

By a similar trick you can get a discontinuous modification of Brownian motion, with which I think you can get an example where $\hat{\tau}$ is a.s. finite.

I think the following example says no.

Consider the state space $\{0,1\}$. Let $U$ be a uniform random variable on $[0,1]$ and let $X_t = 1$ if $t=U$ and $X_t = 0$ otherwise. Note that $X_t$ is a.s. not cadlag, but for each $t$ we have $X_t = 0$ a.s., so $X_t$ is a modification of the cadlag process $\hat{X}_t$ that just sits at 0. Taking $\mathcal{G}_t = \sigma(X_s : s \le t)$, unless I am mistaken, $X_t$ meets the Revuz/Yor definition of a Markov process (under its natural filtration). We may take the transition semigroup to simply be the identity $P_t f = f$, which is Feller.

But if we let $\tau = \inf\{t : X_t \ne 0\}$ and let $f(0)=0$, $f(1)=1$, then $E[f(X_\tau)] = 1$ while since $\hat{\tau} = +\infty$ we might consider $E[f(\hat{X}_{\hat{\tau}})] = 0$.

By a similar trick you can get a discontinuous modification of Brownian motion, with which I think you can get an example where $\hat{\tau}$ is a.s. finite.