Let $ G $ be a completely reducible algebraic group (group for which every finite dimensional algebraic representation is completely reducible). Let $ H $ be an algebraic subgroup of $ G $ (then $ H $ must also be completely reducible).
Let $ V $ be a finite dimensional representation of $ G $. In general there may be $ H $ subreps of $ V $ which are not $ G $ subreps. For example take $ G=GL(2,\mathbb{C}) $, $ V=\mathbb{C}^2 $ the natural representation and $ H $ trivial. Then any 1d subspace is an $ H $ subrep but not a $ G $ subrep.
However, what is we furthermore suppose that every irreducible $ G $ subrep of $ V $ is also irreducible as an $ H $ subrep of $ V $. Does that imply that every $ H $ subrep of $ V $ is also a $ G $ subrep of $ V $?
No. For a simple example, let $G$ be any group with a nontrivial $1$-dimensional representation and $H$ be trivial. If you take a direct sum $V=V_1\oplus V_2$ of two non-isomorphic $1$-dimensional representations of $G$, then each irreducible subrepresentation of $V$ is still irreducible over $H$. However, there are subrepresentations over $H$ that are not subrepresentations over $G$ (take any $1$-dimensional subspace of $V$ different from $V_1$ and $V_2$).