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.

<B>Edit.</B>  As Sasha points out, it is unnecessary to pullback to $G$ (it just makes it easier for me to think about).