A strong version of the loop theorem implies that if an essential closed curve on the boundary of a 3-manifold $M$ is nullhomotopic, then realizing its image in $\partial M$ as a 4-valent graph, you can draw a cycle with no edge repeats that is homotopic on $\partial M$ to the boundary of an essential embedded disk in $M$. (Compare with Bing's "The geometric topology of 3-manifolds", pg 205.)


Q: Is there a similarly strong version of the annulus theorem?


For instance, suppose you have a (say, hyperbolizable) 3-manifold M with incompressible boundary, and you have two closed curves on the boundary of M that bound an essential singular annulus in M. (E.g., they're homotopic in M but not within its boundary.) Homotoping them to be in general position, can you find two cycles with no edge repeats on the 4-valent graph that is the union of their images, such that the two corresponding curves are homotopic to the two boundary components of an essential embedded annulus in M?


The closest thing I've found is Thm 2 in Cannon-Feustel's "Essential embeddings of annuli...", but that requires the two original curves to be disjoint.



A strong annulus theorem for 3-manifolds
I think this might follow from JSJ theory. Assume that $M$ is irreducible with incompressible boundary. Then any essential annulus is homotopic into an $I$-bundle region or a Seifert-fibered region of the JSJ decomposition. In the Seifert case, the region meets the boundary in annuli, in which case the boundaries of the immersed annulus will be a multiple of the core of the annuli, and I think one can cut and paste to get a simple closed essential core of each annulus, which then cobound an embedded essential annulus. Otherwise, if the annulis lies in a product region, then your immersed annulus goes between two curves on the surface. So this amounts to asking: for a given homotopy class of curve on a surface, can one always cut and paste a subset of the curve to give an embedded curve in the same isotopy class? This seems plausible, e.g., if the curve is not filling (one should be able to cut and paste to get a boundary component of the subsurface that it fills). In any case, it seems that at least this gives a reformulation of the problem (although I'm not sure what happens for twisted $I$-bundles). 


