If the boundary point is not an isolated singularity, then you can't say it is an essential singularity, so the answer to your final question is no. It is quite possible that $f$ cannot be extended to a larger domain that contains a punctured neighborhood of $x$. To be quasi-explicit, take a holomorphic function whose domain of holomorphy is $\Omega$ and multiply by a holomorphic function defined everywhere in the plane except with an essential singularity at $x$.
If $f$ has an isolated singularity at $x$, then yes, this characterizes essential singularities. $f$ goes to $\infty$ at a pole, so in particular is nonzero in a neighborhood of the pole. If the singularity at $x$ is removable, then either $f$ goes to a nonzero value, in which case it is nonzero in a neighborhood of $x$, or $f$ goes to 0. In the latter case, $f$ could be extended to a holomorphic function on a domain containing $x$, so if $x$ were a limit point of the zero set of $f$, then $f$ would be identically zero. And you could easily adjust this to nonzero cases.
You don't need anything near the strength of Picard's theorem. In fact, that is what is so amazing about Picard's theorem: Either $f$ has really nice behavior near an isolated singularity, or it maps everywhere except possibly one point in each neighborhood of the singularity.