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.

This is not exactly what is asked for, rather it is a different proof of the fact in question.

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.

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 relevanat papers by Kov'acs and . Chatzistamatiou and K. R"ulling). 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.

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.