No, that is not true, for essentially the same reason that I explained a couple of days ago in a different question. For simplicity, assume that $R$ contains its residue field $k$. Let $(X,x)$ be a smooth, projective curve over $k$ of genus $g>0$ together with a $k$-point, so that $\text{Pic}^0_{X/k}$ is a smooth, projective variety of positive dimension $g$, and there is a Poincaré sheaf $\mathcal{L}$ on $X\times_k \text{Pic}^0_{X/k}$ whose restriction to $\{x\}\times \text{Pic}^0_{X/k}$ is isomorphic to the structure sheaf.

Let $v:\text{Spec}(\widehat{R})\to \text{Pic}^0(X)$ be a $k$-morphism that does not factor through $\text{Spec}(\widehat{R}) \to \text{Spec}(R)$. For instance, let $k$ be characteristic $0$, let $u:\text{Pic}^0_{X/k} \to \mathbb{A}^g_k$ be a morphism that is unramified at some point $[L]$ mapping to $0$, let $R$ be the Henselization of $k[t]_{\langle t \rangle}$, and let $v$ be the morphism sending the closed point to $[L]$ and with $u\circ v$ sending $t$ to $(e^t-1, 0, \dots, 0)$. For such a morphism $v$, the pullback of $\mathcal{L}$ to $X\times_k \text{Spec}(\widehat{R})$ restricts on every $X\times_k \text{Spec}(R/\mathfrak{m}^n)$ to an invertible sheaf $\mathcal{L}_n$, and there is no algebraization over $\text{Spec}(R)\times_k X$.