$\newcommand\card[1]{\lvert#1\rvert}$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, 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, 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 to Number of points on a linear algebraic group over a finite field.) 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, Proposition 2.1.12(1), the unipotent radical of an opposite parabolic to $H$, which has at least $q$ rational points, injects into $G/H$.

As you have pointed out, [Borel - Linear algebraic groups, Corollary 16.5(ii)], allows us to identify the set of rational points in a quotient (by a connected group) with the quotient of the groups of rational points. We do so freely.

$\newcommand\card[1]{\lvert#1\rvert}$Suppose first 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, 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, $\card{G(\mathbb F_q)/H(\mathbb F_q)}$ equals $\card{(T_G/T_H)(\mathbb F_q)}q^{\dim(G/T_G) - \dim(H/T_H)}$. 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 to Number of points on a linear algebraic group over a finite field.) Otherwise, $\dim(G/T_G) - \dim(H/T_H)$ is positive.

Now drop the assumption that $G$ is solvable. 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)], with two caveats. First, the statement there requires 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. Second, the result there seems only to guarantee that we may choose a Borel subgroup of $G_{\overline{\mathbb F_q}}$ containing $(B_H)_{\overline{\mathbb F_q}}$; but examining the proof shows that we may take $B_G$ to be any Borel subgroup of $G$ containing $B_H$, i.e., any point of the non-empty set of rational fixed points of $B_H$ on the flag variety of $G$.) 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, Proposition 2.1.12(1)], the unipotent radical of an opposite parabolic to $H$, which has at least $q$ rational points, injects into $G/H$.

$\newcommand\card[1]{\lvert#1\rvert}$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 $G/H$.

$\newcommand\card[1]{\lvert#1\rvert}$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, 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, 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 to Number of points on a linear algebraic group over a finite field.) 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, Proposition 2.1.12(1), the unipotent radical of an opposite parabolic to $H$, which has at least $q$ rational points, injects into $G/H$.