Let $X$ be a Noetherian scheme, $x \in X$ a closed point. Denote by $\hat X$ the completion of $X$ along $x$. Now assume that two coherent modules $F, G$ on $X$ coincide over $\hat X$, i.e. $i^*F = i^*G$ for the natural morphism $i: \hat X \to X$. When do we have $F = G$ in a neighborhood of $x$?
My intuition says to me that $\hat X$ is something like a neighborhood of $x$, but of course it should be considered only as a formal neighborhood. So probably, if there is some smoothness assumption on $X$ around $x$, then we may drop formally, i.e. each formal neighborhood gives us a "real" one?
I would appreciate any hint, explanation, reference or an example.
EDIT: Regarding the comments the equalities $i^*F = i^* G$ and $F=G$ should/could be considered as $i^*F \cong i^* G$ and $F \cong G$.
I realised after writing this that I had misread your question - I thought you wanted $i^*F$ and $i^*G$ to be isomorphic, but you actually said they should be equal. Still, here are my thoughts.
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A formal neighborhood is definitely smaller than a Zariski-open neighborhood. It can be helpful to think of $\hat{X}$ as the union or direct limit of all infinitesimal neighborhoods of $x$ in $X$, where "infinitesimal neighborhood" means a subscheme $Y \subset X$, probably nonreduced, that is only supported at $x$. (This fits with the inverse limit on the algebraic side, since Spec is contravariant.)
A nice example showing the difference: let $X = \text{Spec}\ k[x,y]$, $\hat{X} = k[[x,y]]$, the completion of $X$ at the origin. Let $\mathcal{F}, \mathcal{G}$ be, respectively, the structure sheaves $k[x,y] / (x^2 - y^2)$ and $k[x,y] / (y^2 - x^2(x+1))$. These are, respectively, the union of the lines $y = \pm x$, and the nodal cubic with node at the origin, having tangent lines $y = \pm x$.
In this case, $i^*\mathcal{F}$ and $i^* \mathcal{G}$ are actually isomorphic, but not equal, on any infinitesimal neighborhood of the origin, since both curves look infinitesimally like two arcs meeting at a point. You can see this isomorphism on the level of power series: $\frac{k[[x,y]]}{(y-x)(y+x)} \cong \frac{k[[x,y]]}{(y-x\sqrt{x+1})(y+x\sqrt{x+1})}$, using the Taylor series for $\sqrt{x+1}$, which is a unit in $k[[x,y]]$.
Of course, the two curves are neither equal nor isomorphic on any open neighborhood of the origin, since their local rings are different (the first is not a domain, but the second is.)
Anyway, you asked about equality, and in the above example, $i^*\mathcal{F}$ is not equal to $i^*\mathcal{G}$, since the arcs are actually different arcs (except for the first-order neighborhood, since they have the same tangent lines).
I'm not totally sure, but I think the answer to your question is yes and follows from something like the theorem on formal functions (Hartshorne III.11.1), though note that it has no smoothness assumptions. I think you want to consider the sheaf $\mathcal{R} = \mathcal{Hom(F,G)}$. The theorem says that $\mathcal{R} \otimes k[[x,y]]$ is the inverse limit of the restrictions of $\mathcal{R}$ to infinitesimal neighborhoods. If $i^*F = i^*G$ on $\hat{X}$, then there's an "identity map" $\hat{\varphi} \in \mathcal{R} \otimes k[[x,y]]$. So certainly $\mathcal{R}$ can't be too small (e.g. it can't be 0), and we can write
$\hat{\varphi} = \varphi_1 \otimes f_1 + \cdots + \varphi_n \otimes f_n,$
for some maps $\varphi_i \in \mathcal{R}$ and power series $f_i \in k[[x,y]]$. I'm not sure how to complete this argument, I'm afraid.
SECOND EDIT:
Faithful flatness is what I was forgetting - never mind the theorem on formal functions. The answer to the "equality" version of the question is yes: following Cantlog's suggestion in the comments, suppose $F$ and $G$ are comparable, say $F$ is a subsheaf of $G$, and $i^*F = i^*G$. Then $i^*(G/F)$ is zero, and by faithful flatness of $\hat{O}_{x,x}$ over $O_{x,X}$, this forces $G/F \otimes O_{x,X}$ to be zero (where $O_{x,X}$ is the local ring of $x \in X$), so $F = G$ over the local ring. Hence they are equal on an open neighborhood, since everything is finitely-generated.
Hopefully that completes the answer to your question!