$\newcommand\card[1]{\lvert#1\rvert}$Suppose that $G$ is solvable, and let $T_G$ be a maximal torus.  Then the inclusion of $T_G$ in the reductive quotient of $G$ is an isomorphism, by, say, [Milne - Algebraic groups](https://doi.org/10.1017/9781316711736), Theorem 16.33.  Since the unipotent radical of $G$ is split, by [Milne], Corollary 16.24, we have that $\card{G(\mathbb F_q)}$ equals $\card{T_G(\mathbb F_q)}q^{\dim(G/T_G)}$.  An analogous computation holds for $H$, so, with the obvious notation (and using [Borel - Linear algebraic groups](https://doi.org/10.1007/978-1-4612-0941-6), Corollary 16.5(ii), as you have already noticed, to observe that $T_G(\mathbb F_q)/T_H(\mathbb F_q) \to (T_G/T_H)(\mathbb F_q)$ is a bijection), $\card{G(\mathbb F_q)/H(\mathbb F_q)}$ is at least $\card{(T_G/T_H)(\mathbb F_q)}q^{\dim(G/T_G) - \dim(H/T_H)}$.  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.  (See [my answer](https://mathoverflow.net/a/444877) to [Number of points on a linear algebraic group over a finite field](https://mathoverflow.net/questions/444855).)  Otherwise, $\dim(G/T_G) - \dim(H/T_H)$ is positive.

In general, there is a Borel subgroup $B_G$ of $G$ such that $(B_G \cap H)_\text{red}$ is a Borel subgroup $B_H$ of $H$.  (This is stated in [Borel], Proposition 11.14(2), after passing to the identity component; but $(B_G \cap H)_\text{red}$ normalizes the parabolic subgroup $(B_G \cap H)_\text{red}^\circ = B_H$, hence is contained in it, so actually we don't need to pass to identity components.)  Thus the natural map from $B_G(\mathbb F_q)/B_H(\mathbb F_q)$ to $G(\mathbb F_q)/H(\mathbb F_q)$ is an injection.  If $B_H$ doesn't equal $B_G$, then $\card{B_G(\mathbb F_q)/B_H(\mathbb F_q)}$, hence $\card{G(\mathbb F_q)/H(\mathbb F_q)}$, 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 $G$.  By, for example, [Conrad, Gabber, and Prasad - Pseudo-reductive groups](https://doi.org/10.1017/CBO9781316092439), Proposition 2.1.12(1), the unipotent radical of an opposite parabolic to $H$, which has at least $q$ rational points, injects into $G/H$.