So, for those, who don't know all the statements of the cited paper by heart, the mentioned corollary states that in the situation described the following vanishing holds:
$$R^{n-1}\pi_*\mathcal O_Y(-E)=0.$$

This is almost automatic for rational singularities of surfaces by your suggested approach. In that case you actually do not get the isomorphism that you claim but that $R^{n-1}\pi_*\mathcal O_Y(-E)\simeq \mathrm{coker}[\pi_*\mathscr O_Y \to \pi_*\mathscr O_E]$, but that's why it works, since the latter is indeed $0$.

As Karl points out in his comment it is also pretty easy for isolated rational singularities in arbitrary dimension. The essence of the argument is that if the singularity is a deformation retract of a neighbourhood, then the restriction map on (singular) cohomology from $Y$ to $E$ is surjective. Using Hodge theory this means that the same is true for the natural morphism
$$
R^i\pi_*\mathscr O_Y \to R^i\pi_*\mathscr O_E
$$
and you get the vanishing you need for your argument to work.
For more details see Lemma 2.14 in *Mixed Hodge structures associated with isolated singularities* by Steenbrink.

One should note that here we are actually using the fact that normal crossings singularities are Du Bois (in the Hodge theoretic part), but I accept that this is indeed simpler than using Du Bois singularities in general.

In higher dimensions in general I am not sure how to make this work. Perhaps by some clever use of general hyperplane sections you can prove that the support of this sheaf is zero-dimensional, but I don't know how to do more.

There is a good reason we used Du Bois singularities in the paper and instead of trying to avoid them, you should perhaps embrace them. ;)

In higher dimensions ($n>2$) in order to prove what you want you would need $R^{n-2}\pi_*\mathcal O_E$ to vanish.
The $n-2$ here doesn't seem to allow a special treatment (unlike the $n-1$) this is probably as hard as proving this for all $i>0$ (or at least $i>n-3$ which in dimension $3$ is the same thing).
Given that you assume that $X$ has rational singularities this is equivalent to the natural morphism
$$
R^i\pi_*\mathscr O_Y \to R^i\pi_*\mathscr O_E
$$
being an isomorphism.
Now, then if $X$ has only isolated singularities this is equivalent to having Du Bois singularities.

Of course, we actually have a simple proof for isolated singularities, but my feeling is that the truth is that this vanishing really works because of the Du Bois property, which in the isolated case can be grasped easily and so in that case we can seemingly avoid it, but it is there anyway.