My question is whether any compact smooth surface in $\mathbb R^3$ (with smooth boundary) can be completed to a closed smooth surface in $\mathbb R^3$ without boundary? It is easy to complete it to an abstract smooth closed $2$-manifold by gluing disks to boundary circles. It seems, however, that in some cases this procedure can't be applied in $\mathbb R^3$ but in all examples that I can imagine we can glue together two (for example) different boundary circles by using a cylinder to obtain a closed surface. Is it a general case or there is an example of a smooth compact surface with smooth boundary in $\mathbb R^3$ which can't be completed to a closed surface in $\mathbb R^3$?
Edit:Lee Mosher in his answer showed that if the surface is not orientable, then it is easy to find a counterexample. What about orientable surfaces?