$\newcommand\card[1]{\lvert#1\rvert}$If $H$ and $G$ are solvable, then $\card{G_\sigma}$ equals $\card{(T_G)_\sigma}q^{\dim(G/T_G)}$, where $T_G$ is a maximal torus in $G$, and analogously for $H$, so $\card{G_\sigma/H_\sigma}$ is at least $\card{(T_G)_\sigma/(T_H)_\sigma}q^{\dim(G/T_G) - \dim(H/T_H)}$. If $T_H$ is a proper subgroup of $T_G$, then $T_G/T_H$ is a torus of dimension at least $1$, hence has at least $q - 1$ points. Otherwise, $\dim(G/T_G) - \dim(H/T_H)$ is positive.
In general, there is a Borel subgroup $B_G$ of $G$ such that $B_G \cap H$ is a Borel subgroup $B_H$ of $H$. If $B_H$ doesn't equal $B_G$, then $\card{G_\sigma/H_\sigma}$ is at least $\card{(B_G)_\sigma/(B_H)_\sigma}$, which is at least $q - 1$. Otherwise, $H$ is a proper parabolic subgroup of $G$, and we may, and do, replace $H$ and $G$ by their images in the reductive quotient of $G$. Then the unipotent radical of an opposite parabolic to $H$, which has at least $q$ rational points, injects into $G/H$.