For a counterexample take $p=1$, $n=d=2$, $P_1(z_1, z_2) = 2z_1z_2$, and $Q_1(z_1, z_2) = z_1^2 + z_2^2$. Then $H(z_1, z_2) \leq 0$ for all $z_1, z_2\in\mathbb{R}$, but $H(1, i) = 4$.

For $d=1$ this is true because $H(z)$ is a bilinear form on $\mathbb{R}^n$ and nonpositive definite by assumption, so it is of the form $H(z) = -\lVert Az\rVert^2$ for some real matrix $A$, and therefore nonpositive definite on $\mathbb{C}^n$ as well. For a counterexample with $d=2$ it is enough to consider two variables and $p=1$ with $P_1(z_1, z_2) = 2z_1z_2$ and $Q_1(z_1, z_2) = z_1^2 + z_2^2$. Then $H(z_1, z_2) \leq 0$ for all $z_1, z_2\in\mathbb{R}$, but $H(1, i) = 4$.