No, not really. In dimension 4, for example, an open subset of $\mathbb{R}^4$ can be homeomorphic to $\mathbb{R}^4$ but not diffeomorphic, as there are exotic smooth $\mathbb{R}^4$'s that embed smoothly in $\mathbb{R}^4$.
But in dimensions different from 4, $\mathbb{R}^n$ admits a unique smooth structure. So your neccessary and sufficient condition can be that the open subset is homeomorphic to $\mathbb{R}^n$. That's probably not what you want to hear?
