I suspect that you will have trouble with trying to "simultaneously" resolve $X$ and $X/G$. We actually run into this issue in the joint paper with Anatoly Libgober arXiv:math/0206241 (although it may not be evident from the paper) and consequently had to settle for working with $\hat X\to\hat Y$ which was $G$-equivariant morphism of smooth varieties which is not necessarily a quotient map. It didn't matter for us a whole lot, since we were working with the formalism of log-terminal pairs.
However, it may not be completely impossible to find this "simultaneous resolution", especially if you know something about $X$ (say it is of small dimension). You can start by using abelianization trick of Batyrev (arXiv:math.AG/0009043) to reduce to the situation when the stabilizers of the action of $G$ are abelian. But then you get into a bit of trouble.
Let me illustrate this trouble by the simplest example, that of $A_2$ singularity. Let $G=\mathbb Z_3$ act on $X=\mathbb C^2$ by $(x,y)\mapsto (xw,yw^{-1})$ where $w=\exp(2\pi I/3)$.
One may view the quotient in terms of toric geometry: one starts with the toric variety that corresponds to the cone $C=(\mathbb Q_{\geq 0})^2$ in the lattice $N=\mathbb Z^2$. Taking the quotient by the above group has the effect of enlarging the lattice $N$ to a coindex three suplattice $N_1=\mathbb Z^2+ {\mathbb Z}(\frac 13,-\frac 13)$. The minimal resolution of singularities of the quotient corresponds to subdividing the cone $C$ into three cones $$ \mathbb Q_{\geq 0}(1,0)+\mathbb Q_{\geq 0}(\frac 23,\frac 13),~~ \mathbb Q_{\geq 0}(\frac 23,\frac 13)+\mathbb Q_{\geq 0}(\frac 13,\frac 23),~~ \mathbb Q_{\geq 0}(\frac 13,\frac 23)+\mathbb Q_{\geq 0}(0,1) $$ which are generated by lattice elements of $N_1$ (the two additional rays correspond to the two exceptional divisors of the minimal resolution).
Taking the obvious $\hat X$, which is the normalization of resolution of $Y$ in the field of functions of $X$ corresponds to considering the same subdivision, but now in the original lattice $N$. The same cones can be written now in terms of minimum generators as $$ \mathbb Q_{\geq 0}(1,0)+\mathbb Q_{\geq 0}(2,1),~~ \mathbb Q_{\geq 0}(2,1)+\mathbb Q_{\geq 0}(1,2),~~ \mathbb Q_{\geq 0}(1,2)+\mathbb Q_{\geq 0}(0,1). $$ The middle cone is now not generated by a basis of the lattice (determinant is three) and thus gives a singular point.
If one is looking for a toric "simultaneous" resolution of this singularity, then one needs to make a subdivision in such a way that for each of the smaller cones taking minimum generators in either lattice $N$ or $N_1$ gives a basis of the said lattice. This requires that in each cone exactly one of minimum generators of $N$ coincides with that of $N_1$. It is possible to do it in this example by splitting the middle cone into $$ \mathbb Q_{\geq 0}(2,1)+\mathbb Q_{\geq 0}(1,1),~~ \mathbb Q_{\geq 0}(1,1)+\mathbb Q_{\geq 0}(1,2) $$ which becomes $$ \mathbb Q_{\geq 0}(\frac 23,\frac 13)+\mathbb Q_{\geq 0}(1,1),~~ \mathbb Q_{\geq 0}( 1,1)+\mathbb Q_{\geq 0}(\frac 13,\frac 23) $$ in $N_1$.
So in this example such $\hat X$ exists. You can try playing with other surface singularities, say $A_n$ for starters. If I were to bet on this, I would say that if you can handle $A_4$, then you should be able to get all of $A_n$.