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.