Let $K$ be a quadratic number field, and let $E_1$ and $E_2$ be two isogeneous elliptic curves over $K$. Assume we know that $j(E_1)^\sigma=j(E_2)$ where $\sigma$ is the generator of the Galois group of $K/Q$. Can we then say that some twist of $E_1$ is a $Q$curve? If so, is there a good way of describing the necessary twist?
Yes  a Qcurve is one whose geometric isogeny class is preserved by Galois, and that's evidently the case here. Of course there is no guarantee that E_1 and its Galois conjugate are isogenous over K. Is that the question you're asking? If so, I think there's a cohomological criterion for this due to Quer  at least that's what I say in Remark 2.9 of my paper with Chris Skinner about this stuff:

$\begingroup$ I was mostly wondering if there is a twist that makes E_1 a Qcurve over K which, as you point out, is not necessarily so, and the criteria for it is given by Quer. $\endgroup$ – Soroosh Sep 2 '11 at 17:47