The answer is yes. In fact, there is the following

Proposition. Every $2$-dimensional rational singularity $(X, \, x)$ is analytically $\mathbb{Q}$-factorial, i.e. there exists an analytic neighborhood $V$ of $x$ such that every Weil divisor on $V$ is a $\mathbb{Q}$-Cartier divisor.

In particular, every $2$-dimensional rational singularity is $\mathbb{Q}$-Gorenstein.

See S. Ishii, Introduction to Singularities, Theorem 7.3.2.

The answer is yes. In fact, there is the following

Proposition 1. Every $2$-dimensional rational singularity $(X, \, x)$ is analytically $\mathbb{Q}$-factorial, i.e. there exists an analytic neighborhood $V$ of $x$ such that every Weil divisor on $V$ is a $\mathbb{Q}$-Cartier divisor.

In particular, every $2$-dimensional rational singularity is $\mathbb{Q}$-Gorenstein.

Concerning the particular case of rational Gorenstein singularities, we have

Proposition 2. For a $2$-dimensional normal singularity $(X, \, x)$, the following are equivalent:

1. $(X, \, x)$ is a rational double point (i.e., a Du Val singularity);
2. $(X, \, x)$ is a rational hypersurface singularity;
3. $(X, \, x)$ is a rational Gorenstein singularity;
4. $(X, \, x)$ is a canonical singularity.

See S. Ishii, Introduction to Singularities, Theorem 7.3.2 and Theorem 7.5.1.

The answer is yes. In fact, there is the following

Proposition. Every $2$-dimensional rational singularity is analytically $\mathbb{Q}$-factorial, in particular $\mathbb{Q}$-Gorenstein.

See S. Ishii, Introduction to Singularities, Theorem 7.3.2.

The answer is yes. In fact, there is the following

Proposition. Every $2$-dimensional rational singularity $(X, \, x)$ is analytically $\mathbb{Q}$-factorial, i.e. there exists an analytic neighborhood $V$ of $x$ such that every Weil divisor on $V$ is a $\mathbb{Q}$-Cartier divisor.

In particular, every $2$-dimensional rational singularity is $\mathbb{Q}$-Gorenstein.

See S. Ishii, Introduction to Singularities, Theorem 7.3.2.