I suspect it will almost never be surjective. If $K$ is a finitely generated field, the Picard group over $K$ is a finitely generated abelian group (Mordell-Weil), while over the algebraic closure, it's definitely not finitely generated, it has both infinite torsion and "infinite rank". If $K = \mathbb{R}$, then the Picard group is a $g$-dimensional real manifold and, over $\mathbb{C}$ is a $g$-dimensional complex manifold.
My guess is that, for a positive dimensional algebraic variety $V/K$, if $K \ne \bar{K}$, then $V(K) \ne V(\bar{K})$. So this has nothing to do with Picard groups.