Let $R$ be a discrete valuation ring with algebraically closed residue field $k$. Let $K:=\mathrm{Frac}(R)$ the fraction field of $R$. Suppose $K$ is of characteristic zero. Denote by $\overline{K}$ the algebraic closure of $K$ and $\overline{R}$ the integral closure of $R$ in $\overline{K}$. Since $k$ is algebraically closed, the residue field of $\overline{R}$ is $k$.

Let $\pi:\mathcal{X} \to \mathrm{Spec}(R)$ be a flat family of projective varieties and smooth generic fiber $\mathcal{X}_{K}$. Denote by $\overline{\pi}:\mathcal{X}_{\overline{R}} \to \mathrm{Spec}(\overline{R})$ the base change of $\pi$ under the natural inclusion $R \hookrightarrow \overline{R}$. Note that the generic fiber of $\overline{\pi}$ is again smooth and the special fiber is the same as $\mathcal{X}_k$ (as $k$ is algebraically closed). Let $E$ be a coherent sheaf on $\mathcal{X}_k$. My question is:

If $E$ lifts to a coherent sheaf $E_{\overline{R}}$ on $\mathcal{X}_{\overline{R}}$ (in the sense, $E_{\overline{R}}$ is flat over $\overline{R}$ and $E_{\overline{R}} \otimes_{\overline{R}} k \cong E$) does it also lift to a coherent sheaf on $\mathcal{X}$ i.e., does there exists a coherent sheaf $E_R$ on $\mathcal{X}$, flat over $R$ such that $E_{{R}} \otimes_{{R}} k \cong E$? If not true in general, is there any known condition on $R$ or $K$ under which this holds true?

**EDIT** Does the above question have a positive answer, if we substitute coherent by locally free?