Let $\hat{X} = \text{Spf} \hat{A}$ be obtained as the formal completion of an affine scheme $X = \text{Spec} A$ where $A$ is an adic noetherian ring. Given a coherent sheaf $\mathfrak{F}$ on $\hat{X}$, is it always possible to find a coherent sheaf $\mathcal{F}$ on $X$ such that $\hat{\mathcal{F}} = \mathfrak{F}$?

According to EGA I (Springer edition), Theoreme (10.10.2), there is an equivalence of categories between the category of finitely generated $A$modules and the category of coherent $O_{{\rm Spf}(A)}$modules. In particular, the answer to your question is Yes. 


Take $A=k[x,y]$, $\hat A=k[[x,y]]$, and suppose coherent sheaf corresponds to the $\hat A$module $N=\hat A/(f)$, where $f\in k[[x,y]]$ is not "algebraic": say, $f=yexp(x)$, assuming $k$ has characteristic zero. It is clear that $N$ does not come from completion of any f.generated $A$module $M$. (Proof: Suppose $N=\hat M$. Since $N$ has no $A$torsion, any torsion of $M$ maps to zero, so we can replace $M$ with $M/(torsion)$ and assume $M$ is torsionfree. Then $M$ embeds in a locally free module (its second dual), but that would make $N$ embed in the free $\hat A$module, which is false.) 

