No, this is not even true for line bundles.
For an example, let $X = \mathbb{P}^2$ and $Y$ a smooth curve in $X$ of genus $>0$ (over an algebraically closed field). Since $X$ is smooth, any line bundle on an open set $U$ extends to a line bundle on $X$ so the map $Pic(X) \operatorname{Pic}(X) \to Pic(U)$ \operatorname{Pic}(U)$ is surjective. However, Since $Pic(X) \operatorname{Pic}(X) \cong \mathbb{Z}$ whereas mathbb{Z}$, it follows that the image of $Pic(Y)$ \operatorname{Pic}(U)$ in $\operatorname{Pic}(Y)$ is of rank $1$ and is independent of $U \supset Y$. Since $\operatorname{Pic}(Y)$ is not even finitely generated , so the map we see that there exist (many) line bundles on $Pic(U) \to Pic(Y)$ is Y$ which do not surjectiveextend to any open $U \supset Y$.
