Yes. By Chevalley-Shepard-Todd, $S(V)^G$ and $S(V)^H$ are polynomial rings. Let $S(V)^G=\mathbb{R}[g_1, \ldots, g_n]$ and $S(V)^H = \mathbb{R}[h_1,\ldots, h_n]$ where the $g_i$ and $h_i$ are homogenous. Then $$S(V)_G^H = \mathbb{R}[h_1,\ldots,h_n]/\langle g_1,\ldots, g_n \rangle.$$ Here the denominator is the ideal of $\mathbb{R}[h_1,\ldots,h_n]$ generated by the $g_i$, and $g_i \in \mathbb{R}[h_1,\ldots,h_n]$ because $S(V)^G \subseteq S(V)^H$.
So $S(V)^H_G$ is a complete intersection, and therefore Gorenstein. Saying $R$ is a finite dimensional Gorenstein $k$ algebra (for $k$ a field) exactly means that there is a linear functional $\int: R \to k$ such that $\langle f,g \rangle = \int fg$ is a perfect pairing. If $R$ is graded, then this functional is "take the top degree piece". See Eisenbud, Commutative Algebra with a view toward Algebraic Geometry Chapter 21.2 for a good overview of finite dimensional Gorenstein algebras.