I'm glad to have stumbled on your question -- I'm actually working on something along the lines of what you're asking about now with Eric Larson. If what we think we've proven is true (and don't quote this yet, since we're not yet even done with the write-up,) then in fact you can say something stronger: you cannot have isogenies of order p for any prime p sufficiently large (where "sufficiently large" depends on K), as long as K does not contain the class field of an imaginary quadratic extension in which p splits (this exactly gets rid of primes which are isogenies of a CM curve defined and with CM over K). So this would mean that if K is quadratic imaginary, there can be no p-isogenies for *any* p sufficiently large unless K is one of the finitely many quadratic imaginary fields of class number one (in which case, if our arguments work, this would mean that all but finitely many possible prime degrees of isogeny p split over K.) 

If you're interested in what happens to primes that split, you can ask whether, more generally, for *any* number field K, there are only finitely many primes that can be isogenies for a curve over K without CM. This seems harder to prove, and as far as we know, is an open problem (to prove it, you'd need to have a good way of distinguishing CM from non-CM curves, or carefully counting points on $X_0(N)$). But hypothetically this is also true: it fits into the general framework of Serre's hypothesis that "exceptional primes" of curves without CM are bounded by a constant depending only on $K$.