I'll think more about this, but if the curvature is positive the necessary and sufficient conditions for the curve to be closed and without self intersections is that the total curvature equal $2 \pi$ and that if you set $\tilde{\kappa}(\theta)$ to be equal to the value of $\kappa(s)$ whenever $\gamma'(s)$ is perpendicular to $(\cos(\theta),\sin(\theta))$ and the determinant of the pair of vectors is positive (we want the oriented normal), then $$ \int_0^{2\pi} e^{-i \theta} \tilde{\kappa}(\theta)\, d\theta = 0. $$
This is because in that case you can solve the differential equation $h''(\theta) + h(\theta) = \tilde{\kappa}(\theta)$ and $h$ will be the support function of an oval. It's not terribly explicit ...