It is even trivial outside appropriate hyperplane section. Let $V$ be a vector space of dimension $n$ such that the Grassmannian is $Gr(r,V)$, and let $V_0 \subset V$ be a subspace of codimension $r$. Consider the composition
$$
f: S \to V \otimes O \to (V/V_0) \otimes O.
$$
It is an isomorphism outside of the zero locus of the determinant
$$
\det(f): \det(S) \to \det(V/V_0) \otimes O.
$$
It remains to note that $\det(S) \cong O(-1)$, so the vanishing locus of $\det(f)$ is a hyperplane section. To be more precise, it is the Schubert cycle, parameterizing subspaces of $V$ that intersect $V_0$ nontrivially.

EDIT. If you want $S$ to be trivial on a complement of a general irreducible hypersurface of degree $d > 1$, then this does not hold. Indeed, if $S$ would be trivial, then $\det(S) \cong O(-1)$ would be trivial as well. But the Picard group of a complement of an irreducible hypersurface of degree $d$ in $Gr(r,n)$ is isomorphic to $\mathbb{Z}/d\mathbb{Z}$, and $O(-1)$ is a nontrivial element there.

allhypersurfaces from some degree on or just there exists one? The latter is trivial if you admit reducible hypersurfaces. $\endgroup$