I will explain how to check it in any particular case, but I am not sure whether there is a good classification available (perhaps, it is even easy and I did not think enough about it). Helgason's book and papers of Joe Wolf may contain some relevant information.
Since the complex structure on $M=G/K$ is homogeneous, the image of $H$ is a complex subvariety if and only if its tangent subspace at $e$ is a complex subspace of $T_e M,$ i.e. invariant under the complex structure operator $J: T_e M\to T_e M.$ Let $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ be the Cartan decomposition, then $T_e (G/K)$ may be identified with $\mathfrak{p}$ via the projection onto the second summand and $J=\text{ad}(X_0)$ for a certain element $X_0$ in the center of $\mathfrak{k}$ (if $G$ is simple then this center is one-dimensional). Thus the answer is "yes" if and only if
$$T_e(H/K\cap H)=\mathfrak{h}/\mathfrak{k}\cap\mathfrak{h} \subset \mathfrak{p} \text{ is } \text{ad}(X_0)\text{-invariant.}$$
Subalgebras $\mathfrak{h}$ with this property would need to be classified modulo conjugation by $K$. One can proceed a bit further by complexifying $\mathfrak{g}$, so that $\mathfrak{p}_\mathbb{C}=\mathfrak{p}_{+}\oplus\mathfrak{p}_{-}$ is the eigenspace decomposition for $J_{\mathbb{C}}$ (the eigenvalues are $\pm i$), keeping in mind that a complex Lie subalgebra of $\mathfrak{g}_\mathbb{C}=\mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C}$ is a complexification of a Lie subalgebra of $\mathfrak{g}$ if and only if it is invariant under the complex conjugation.
A final elementary observation is that there always exist nontrivial such subgroups $H,$ for example $NA$ from the Iwasawa decomposition $G=NAK$ projects onto $G/K.$