If you take Riemannian generic manifold of dimension $\ge3$ then convex hull is not closed. BUT it is hard to make explicit and generic at the same time :)
To see it do the following exercise first: Show that in generic 3-dimensional manifold, arbitrary smooth convex surface contains no geodesic. (Here geodesic = geodesic in ambient space.)
To make word "generic" more clear: show that any metric admits $C^\infty$-perturbation such that above property holds.
Hint: Use Jacobi fields to show the following: If geodesic $\gamma$ lies in a convex surface then curvature tensor along $\gamma$ is very speacial.