Whoops. Misread the question....

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