From the comments, it sounds like there might exist a reference for a positive answer to this question. However, I don't know it, so here is an argument using Borel-Tits.
Claim. The maximal smooth connected trigonalizable $K$-subgroups of $G$ are of the form $S \cdot U_G(\lambda)$, where $S$ is a maximal split $K$-torus of $G$ and $\lambda\colon \mathbf{G}_m \to S$ is a regular cocharacter, i.e., $\langle\lambda, \alpha\rangle \neq 0$ for all roots $\alpha$ of $(G, S)$ (and $U_G(\lambda)$ is the notation from the dynamic method).
Proof. First, it is well-known that $S \cdot U_G(\lambda)$ is trigonalizable. By Theorem 8.2 of Borel-Tits, all maximal trigonalizable $K$-subgroups of $G$ are $G(K)$-conjugate, so it's enough to show that $S \cdot U_G(\lambda)$ is maximal. The multiplication map $U_G(\lambda^{-1}) \times Z_G(S) \times U_G(\lambda) \to G$ is an open embedding, so for dimension reasons we need only show that if $V$ is a trigonalizable $K$-subgroup of $G$ containing $S \cdot U_G(\lambda)$ then $(V \cap Z_G(S))/S$ and $V \cap U_G(\lambda^{-1})$ are both finite.
Since $Z_G(S)/S$ is $K$-anisotropic, $S$ is the maximal smooth connected trigonalizable $K$-subgroup of $Z_G(S)$ and thus $(V \cap Z_G(S))/S$ is finite. Moreover, if $V \cap U_G(\lambda^{-1})$ is nontrivial then the Lie algebra of $V$ contains a pair of vectors with opposite $S$-weights, and this cannot happen for a trigonalizable $K$-subgroup of $G$ (reduce to $K = \overline{K}$ and think about a Borel). QED
The question is not sensitive to isogenies, so we may assume $G = Z \times G'$, where $Z$ is a split torus and $G'$ is semisimple and simply connected. By the Claim and Proposition 9.3 of Borel-Tits, $H$ must contain $S \cdot U_G(\lambda)$ for some $S$ and $\lambda$. In particular, $H$ contains $Z$, so we reduce to the case $G = G'$. Now $H$ must also contain $S \cdot U_G(\lambda^{-1})$: indeed, because $H$ is reductive, $\dim U_H(\lambda) = \dim U_H(\lambda^{-1})$. Since also $\dim U_G(\lambda) = \dim U_G(\lambda^{-1})$ and $U_G(\lambda) = U_H(\lambda)$, we get the claim by a dimension argument. (This is the only place where reductivity of $H$ is used. It probably only requires that $H(K)$ be unimodular, as YCor suggested, but I haven't checked carefully.)
Thus the problem reduces to showing that $G$ is generated by $U_G(\lambda)$ and $U_G(\lambda^{-1})$. Note that $G = \prod_{i=1}^n \mathrm{R}_{K_i/K}(G_i)$ for finite separable field extensions $K_i/K$ and absolutely simple $K_i$-groups $G_i$, and $U_G(\lambda^{\pm 1})$ decomposes into a product of subgroups of the $\mathrm{R}_{K_i/K}(G_i)$, so we may assume $n = 1$ (and of course each $G_i$ is isotropic since you are assuming $G$ is totally isotropic). We have that $S$ is the maximal $K$-split torus of $\mathrm{R}_{K_1/K}(S_1)$ for a unique maximal $K_1$-split torus $S_1$ of $G_1$. If $\lambda_1\colon \mathbf{G}_m \to S_1$ is the cocharacter corresponding to $\lambda$, then we have $U_G(\lambda^{\pm 1}) = \mathrm{R}_{K_1/K}(U_{G_1}(\lambda_1^{\pm 1}))$ (just think about weights on the Lie algebra), so it is enough to show that $G_1$ is generated by $U_{G_1}(\lambda_1^{\pm 1})$; in other words, we may assume that $G_1$ is absolutely simple.
The question of whether $G$ is generated by $U_G(\lambda^{\pm 1})$ is purely algebraic, so by passing to $\overline{K}$ it would in fact be enough to show that if $G$ is split simple and $\lambda$ is any nontrivial cocharacter then $G$ is generated by $U_G(\lambda^{\pm 1})$. If $\lambda$ factors through a (split) maximal $K$-torus $T$, then it is enough to show that $G$ is generated by $T$ and $U_G(\lambda^{\pm 1})$: indeed, if $G_0$ is the $K$-subgroup generated by $U_G(\lambda^{\pm 1})$, then $G_0$ is normalized by $T$, so if $G = \langle G_0, T \rangle$ then $G_0$ is normal in $G$, and by simplicity $G = G_0$. Thus we have reduced to the following root-theoretic claim: if $(V, \Phi)$ is an irreducible root system and $f\colon V \to \mathbf{R}$ is a nontrivial linear functional, then $\mathbf{Z}\Phi$ is spanned by the set of roots $a$ such that $f(a) \neq 0$.
For this, note that there is a basis $B$ of $(V, \Phi)$ such that $f(b) \geq 0$ for all $b \in B$. Since $f$ is nontrivial, there is some $a \in B$ such that $f(a) > 0$. It's a general fact about root systems (see Bourbaki, Chap. VI, no. 1.6, Corollaire 3) that if $C$ is a connected subgraph of the Dynkin diagram, then $s_C = \sum_{b \in C} b$ is a root. It is easy to see that $\mathbf{Z}\Phi$ is spanned by $s_C$ as $C$ ranges over connected subgraphs of the Dynkin diagram containing $a$ (to get another $b \in B$, induct on the length of a shortest path from $a$ to $b$), and of course $f(s_C) > 0$ for each such $C$. So we're done!
