If your representation $V$ is irreducible, it carries a unique up to scalars non-degenerate "contravariant form", with the property that $E$ and $F$ are adjoint.  If you have chosen a self-dual basis (so the form looks like dot product) then the matrix form of $F$ is therefore the transpose of $E$.  The matrix $H$ is determined by $H=[E,F]$.

In the general case there are of course many "contravariant forms" on a reducible finite dimensional representation $V$, but perhaps there is a natural choice in whatever situation you are interested in.  

On the other hand, if your question is, given a nilpotent matrix $E$, is there a unique way to extend it to an $sl_2$-representation, the answer is no, but almost.  See paragraph 11 of chapter VIII of Bourbaki's "Lie groups and Lie algebras" on $sl_2$-triplets.  The bottom line is that if you want to pin down matrix forms of $F$ and $H$ just from knowledge of the matrix of $E$ you need a bit more structure (b/c you can conjugate $H$ and $F$ by anything in the centralizer of $E$).