Consider a Seifert fiber space. Is it always possible to find a finite cover that is a circle bundle and the preimage of any fiber is a finite union of circles?
If the Seifert fiber space is compact, then this is true, as long as the base orbifold is "good", which means that it has a finitesheeted manifold cover, which is a compact surface. This induces a cover of the Seifert fiber space which is a circle bundle over the surface. If the base orbifold is bad, then no such covering will exist. This can happen for a Seifert fibering of $S^3$ over a football orbifold with distinct orders of torsion points, or over a teardrop orbifold. If the Seifert fiber space is noncompact, then there may be infinitely many exceptional fibers, and the base orbifold might have torsion of arbitrarily large order, so there is no hope of finding a finiteindex cover which is a circle bundle. See the draft of Thurston's book for more information on orbifolds and Seifert fibered spaces. Exercise 5.7.10 is on the Seifert fibering of $S^3$ over bad orbifolds. 


I believe the answer is yes (although I haven't actually checked). Here's the idea: Given a Seifertfibred space you can think of it as being fibred over a $2$orbifold. You can desingularise that $2$orbifold by taking the appropriate branched cover. Pulling back the Seifertfibering gives you a genuine $S^1$bundle. This skirts the issue of whether or not you can desingularize the $2$orbifold by an appropriate cover but I believe it's not hard to show such "bad" 2orbifolds never occur as the base space to a Seifertfibred $3$manifold. Right, they're classified here: http://en.wikipedia.org/wiki/Orbifold and you can compare that to the base orbifolds of Seifertfibred 3manifolds to answer your question. 

