It is right, and not so obvious.

The question of whether or not a Markov process hits
particular sets is usually studied using the concept of capacity.

For a continuous time parameter Markov process taking values in
a general topological state space, this leads to non-trivial problems of measurability.
For instance, for a Borel $A$ there is no guarantee that the set $R(T,A)\in{\cal F}$
where $(\Omega,{\cal F},\Pr)$ is the probability space.
However, under suitable conditions, capacity theory can be used to show that $R(T,A)$ is universally measurable, and hence that $\Pr[R(T,A)]$ makes sense.

Let's assume that the state space and process are "nice";
say, the state space is a locally compact, separable metric space,
and the process has right continuous sample paths.
For fixed $T<\infty$, the formula $\phi(A)=\Pr[R(T,A)]$ defines a Choquet capacity on the
Borel sets $A$. Therefore, $$\phi(A)=\sup(\phi(K): K\subseteq A,\ K\mbox{ compact}).$$

For a compact $K$, define the stopping time $\tau(\omega):=\inf(t\geq 0: X_t(\omega)\in K)$.
Since the sample paths of $(X_t)$ are right continuous and $K$ is
closed, we have $R(T,K) = (X_{\tau\wedge T} \in K).$

Therefore,
$$\Pr[R(T,K)]\leq \mathbb{E}[I_A(X_{\tau\wedge T})]\leq \Pr[R(T,A)].$$

Taking the supremum over compact subsets of $A$ gives
$$ \Pr[R(T,A)]=\sup_{\tau}\ \mathbb{E}[I_A(X_{\tau\wedge T})],$$
which gives your desired result. Letting $T\to\infty$ gives the infinite version.

The result hinges on the fact that, as far as the process goes, the Borel set $A$ can be well
approximated from the inside by compact sets.

You can find more details in Chapter I, Section 10 of Blumenthal and Getoor's
*Markov Processes and Potential Theory*, or in Section 3.3 of Kai Lai Chung's *Lectures from Markov Processes to Brownian Motion*.

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