I assume you are working in characteristic zero with varieties.  I'm also going to assume you want $X$ to be normal (for simplicity, your rational resolutions will all factor through the normalization).  The proof in other characteristics is hard I think (see relevant papers by Kovács and A. Chatzistamatiou and K. Rülling).  The way I prefer to see this is via Grothendieck duality and also utilizing Grauert–Riemenschneider vanishing.  If you want to work with Q-Schemes with dualizing complexes, everything below works by the version of Grauert–Riemenschneider vanishing by Murayama (but let's stick with varieties).

Ok, if $X$ has a rational resolution $f : Y \to X$, then your hypothesis means that
$$
O_X \to R f_* O_Y
$$
is an isomorphism in the derived category.  In particular, by Grothendieck duality, you have that 
$$
R f_* \omega_Y^{\bullet} \to \omega_X^{\bullet}
$$
is an isomorphism in the derived category as well.  But $Y$ is smooth and so Cohen-Macaulay, and in particular $\omega_Y^{\bullet} = \omega_Y[d]$ (ie, it's a shifted sheaf).  Furthermore by Grauert–Riemenschneider vanishing we have that  $R\pi_* \omega_Y[d] = \pi_* \omega_Y[d]$ is a sheaf.  Thus $\omega_X^{\bullet}$ is a sheaf too and $X$ is Cohen–Macaulay.  In particular, we now have that

 1. $X$ is Cohen-Macaulay.
 2. $\pi_* \omega_Y = \omega_X$.

Ok, now if $f' : Y' \to X$ is another resolution, we can find $Y''$ birationally mapping to both $Y$ and $Y'$.  Furthermore we may assume $Y'$ a smooth variety (you can resolve indeterminacies and then resolve the singularities, or take the product $Y \times_X Y'$, take the irreducible component dominating $X$ and resolve that).  

In particular, since we need to show that $O_X \to R \pi_* O_{Y'}$ is a quasi-isomorphism, by Grothendieck duality it suffices to show that $g_* \omega_{Y'} = \omega_X$.  To do this, it suffices to show that both 
$\pi_* \omega_{Y''} = \omega_{Y}$ and $\nu_* \omega_{Y''} = \omega_{Y'}$ (where $\pi$ and $\nu$ are the relevant maps) since then $f_* \pi_* \omega_{Y''} = \omega_X$ and chasing the diagram the other way we get $g_* \omega_{Y'} = g_* \nu_* \omega_{Y''} = \omega_X$.

Ok, so in particular we need to show that any resolution of a smooth variety is a rational resolution.  But to do that, it is a computation (that I can show some details if you want) that $\pi_* \omega_{Y''} = \pi_* \pi^* \omega_Y \otimes O_Y(D)$ where $D$ is an **effective** divisor.  Ie, the relative canonical over a smooth variety is effective (you might have seen this in other places in notes of Ein).  It follows now that $\pi_* \omega_{Y''} = \omega_Y$ which is what we wanted to show.