Neighborhood fingerprint of a graph

The answer is no.

Among Andries E. Brouwer's web pages at TU Eindhoven, the Cages page lists three non-isomorphic graphs on $70$ vertices (they are $(3,10)$-cages, found by O'Keefe & Wong 1980) with the same distance distribution $1+3+6+12+24+20+4$ around each vertex of each graph.

That is, starting from any vertex $v$, there's one vertex at distance $0$ from it, $3$ immediate neighbor vertices at distance $1$, six at exact distance $2$ (thus nine which are distinct from $v$ and at distance at most $2$ from it, and ten, including $v$ itself - that would give your $F(v,2)$ -, which can be reached from it by traversing at most two edges), and so forth.

Further down the same table, there appear four non-isomorphic $(5,5)$-cages, on $30$ vertices each, all with the same distance distribution $1+5+20+4$ around each vertex.

Elsewhere on those pages, you'll also find corrigenda to the classic reference I had already mentioned in comments: Brouwer-Cohen-Neumaier's Distance-Regular Graphs, Springer 1989.

Answer. Arguably the simplest infinite set of counterexamples to "Does the converse hold?" are provided by the Möbius ladder's vis-à-vis the prism graphs. Any pair of such graphs (of equal orders) bear stronger similarity to one another than an equal neighborhood fingerprint: all subgraphs induced by balls of radius $r$ are isomorphic (hence the OP's hypothesis holds for such pairs), unless $r\geq\lfloor\frac{\text{number of spokes}}{2}\rfloor$, i.e., unless the balls are large enough to engulf the whole graph and detect the non-isomorphy.

This answers the question.

Remark. While we are at it: the above example can be seen as a reason why the (in)famous Reconstruction Conjecture is so incredible: the condition that all non-trivial balls are isomorphic seems very strong---at first sight perhaps even stronger than the hypothesis in the Reconstruction Conjecture. In particular, the largest balls almost 'look' like the vertex-deleted subgraphs featuring in the Reconstruction Conjecture; and yet, the conclusion that the graphs would have to be isomorphic is false.