The answer is **yes** when $X$ is simply connected. This can be proven as follows.

$\underline{\textrm{Step 1.}}$ The fundamental group $\pi_1(X/G)$ is finite. More precisely, $\pi_1(X/G)= G/N$, where $N$ is the smallest normal subgroup generated by those elements in $G$ which have fixed points on $X$.

For a proof, see [M. A. Armstrong, *Calculating the fundamental group of an orbit space*, Proceedings of the American Mathematical Society **84** (1982)].

$\underline{\textrm{Step 2.}}$ Since $X/G$ has only quotient singularities, one has $\pi_1(X/G)=\pi_1(\widetilde{X/G})$. This is proven in Theorem 7.8 of [J. Kollar, *Shafarevich maps and plurigenera of algebraic varieties*, Inventiones Mathematicae **113** (1993)]. See also this MathOverflow question:

Comparing fundamental groups of a complex orbifolds and their resolutions.

The fact that $X$ is simply connected is only used in Step 1. Maybe this step can be refined in order to make it working also in the more general case where $X$ has finite fundamental group.