There is another way to view this (and prove the second part) that is probably worth explaining (and mentioning some characteristic $p > 0$ connections). As abx already pointed out, this problem reduces to computing $R^i f_* O_X$. You don't need resolution of singularities to prove that vanishing though.

Characteristic zero

In characteristic zero, by Grauert-Riemenschneider vanishing, we have that $R^i f_* \omega_X = 0$ for all $i > 0$ (even if $Y$ is singular). On the other hand, since the relative canonical divisor of a projective birational map between regular schemes $f : X \to Y$ is always effective (this is true whenever the relative canonical divisor makes sense, including mixed characteristic, see the work of Lipman on pseudo-rational singularities), we have that $f_* \omega_X = \omega_Y$.

In other words, $R f_* \omega_X \simeq \omega_Y$ in the derived category (they are quasi-isomorphic). Hence by Grothendieck duality for $f$ we see that $O_Y \simeq R f_* O_X$ in the derived category as well. This implies that $R^i f_* O_X = 0$ for all $i > 0$.

Characteristic $p > 0$

IF $Y$ is smooth, then the same results also holds in characteristic $p > 0$.
See

http://front.math.ucdavis.edu/0911.3599https://arxiv.org/abs/0911.3599

However, Grauert-Riemenschneider is known to fail in characteristic $p > 0$ for singular $Y$ so you really need $Y$ to at least have mild singularities. See for instance example 3.11 in

http://arxiv.org/pdf/1212.5105.pdf

It is reasonable to perhaps conjecture that if $Y$ has $F$-regular singularities then Grauert-Riemenschnedier vanishing is true and hence $R^i f_* O_X = 0$ for all $i > 0$ and all resolutions of singularities $f : X \to Y$.