Let's assume the characteristic is $0$.

Let $R$ be a normal $N$-graded ring of dimension 2 with homogenous maximal ideal $m$. Then $R$ has rational singularity if and only if the non-negative degrees part of the graded local cohomology module $H_m(R)$ vanish. That is $H_m(R)_{(i)}=0$ for $i\geq 0$. This is due to Watanabe, see Theorem 2.2 in: http://www.ams.org/tran/2003-355-03/S0002-9947-02-03186-0/home.html

It is safe to work in affine situation, so let $S=k[x,y]$ and $R=S^G$. Then $R$ is normal and generated by forms of positive degree. Since $H_{(x,y)}(S)_{(i)}=0$ for $i\geq 0$, it follows that $H_m(R)_{(i)}=0$ for $i\geq 0$ (one can compute local cohomology in $S$ by using a system of parameters which are elements in $R$).

In characteristic $p$ one probably has to use Frobenius. Note that Boutot's theorem fails in this case (I think it is still true for finite group though).

A truly easy proof is probably not easy to find unless one has a truly elementary definition of rationality.

Let's assume the characteristic is $0$.

Let $R$ be a normal $N$-graded ring of dimension 2 with homogenous maximal ideal $m$. Then $R$ has rational singularity if and only if the non-negative degrees part of the graded local cohomology module $H_m(R)$ vanish. That is $H_m(R)_{(i)}=0$ for $i\geq 0$. This is due to Watanabe, see Theorem 2.2 in: http://www.ams.org/tran/2003-355-03/S0002-9947-02-03186-0/home.html

It is safe to work in affine situation, so let $S=k[x,y]$ and $R=S^G$. Then $R$ is normal and generated by forms of positive degree. Since $H_{(x,y)}(S)_{(i)}=0$ for $i\geq 0$, it follows that $H_m(R)_{(i)}=0$ for $i\geq 0$ (one can compute local cohomology in $S$ by using a system of parameters which are elements in $R$).

In characteristic $p$ one probably has to use Frobenius. Note that Boutot's theorem fails in this case (I think it is still true for finite group though).

A truly easy proof is probably not easy to find unless one has a truly elementary definition of rationality.

EDIT: There are other ways to see this:

II) Again, assume $k$ is algebraically closed of characteristic $0$.Let $S=k[[x,y]]$ and $R=S^G$. Then the following 2 facts will suffice (using same notation as above):

1. There are only finitely many indecomposable reflexive modules over $R$. (Proof not hard, they have to be summands of $S$). In particular, the class group of $R$ is finite.

2. Since $R$ is complete, $R$ has rational singularity is equivalent to the class group of $R$ is finite. This is Theorem 17.4 in Lipman paper on rational singularity.

III) Finally, one can quote Prop 5.15 of Kollar-Mori book on birational geometry. It gave the exact statement, but the proof uses general machinery, and probably close to what you already knew.