I assume that the algebraic group $G$ is smooth and connected, and that you are asking about equivariance for the natural action of $G$ on $H$. There is a quotient morphism $q:G\to H$ that is $G$-equivariant and faithfully flat. Thus, to prove that a $G$-equivariant coherent sheaf on $H$ is locally free, it suffices to check that the pullback $G$-equivariant coherent sheaf on $G$ is locally free. Since $G$ is smooth, for every coherent sheaf $\mathcal{F}$ on $G$, there exists a maximal open subscheme $U$ of $G$ on which $\mathcal{F}$ is locally free (possibly zero), and $U$ is dense in $G$. If $\mathcal{F}$ is $G$-equivariant, then $U$ is $G$-invariant. The only dense $G$-invariant open subset of $G$ is all of $G$. Thus every $G$-equivariant coherent sheaf on $G$ is locally free. Therefore every $G$-equivariant coherent sheaf on $H$ is locally free.
Edit. As Sasha points out, it is unnecessary to pullback to $G$ (it just makes it easier for me to think about).