If $X$ is a seminormal surface, then actually this vanishing is equivalent to $X$ being Du Bois:

1) If $X$ is Du Bois, the vanishing is a direct consequence of Cor.6.2 of this paper, but in this case it is actually pretty easy to prove directly, see #2.

2) Let $P\in X$ be the singular point. There is a standard distinguished triangle
$$ \underline\Omega_X^0 \to \mathscr O_P \oplus R\mu_*\mathscr O_{\widetilde X}\to R\mu_*\mathscr O_E\overset{+1}\to.$$

See 4.11 of Du Bois's paper for this.
It is easy to see (or use Lemma 2.1 of this paper) that this implies that there exists a distinguished triangle
$$R\mu_*\mathscr O_{\widetilde X}(-E)\to \underline\Omega_X^0 \to \mathscr O_P\overset{+1}\to$$

Now observe that $h^0(\underline\Omega_X^0)$ is the seminormalization of $\mathscr O_X$ and
the outside terms have no cohomology higher than $h^1$, so the only non-trivial map is
$$R^1\mu_*\mathscr O_{\widetilde X}(-E)\to h^1(\underline\Omega_X^0)$$
This is an isomorphism, because $\mathscr O_P$ has very little cohomology. It follows that $X$ is Du Bois if and only if $R^1\mu_*\mathscr O_{\widetilde X}(-E)=0$. $\quad\square$

**Remark** It is essential in this proof that $X$ is a surface. In higher dimensions it is true that Du Bois implies the desired vanishing, but I don't think the opposite implication holds. See #1 and my second comment below.