If $X$ and $Y$ are both non separable, then there is no continuous one-to-one linear mapping of $L(X,Y)$ into a Hilbert space. This is because in this case $L(X,Y)$ contains a subspace isomorphic to $c_0(\omega_1)$, and an argument of Olagunju (*A Banach space that cannot be made into a BIP space*, Math. Proc. Cambridge Philos. Soc., **63** (1967) pp 949-950) shows that there is no continuous one-to-one linear mapping of $c_0(\omega_1)$ into a Hilbert space.

Whether or not this rules out any "natural" inner product for you presumably depends on what you are looking for; in any case, what I've written above rules out various possibilities in the non separable setting (in particular, the identity mapping on $L(X,Y)$ being operator-norm-to-inner-product continuous).

As for the separable setting (i.e., $X$ and $Y$ both separable), $L(X,Y)$ is isomorphic to a subspace of $L_\infty([0,1])$ via the Hahn-Banach theorem, which admits a continuous one-to-one linear mapping into the Hilbert space $L_2([0,1])$ (i.e., the formal inclusion operator), hence there is a continuous one-to-one linear mapping of $L(X,Y)$ into the separable Hilbert space $L_2([0,1])$. But whether or not a "natural" such mapping exists, I don't know.

I haven't necessarily answered your question, because I restricted attention to having a continuity requirement, but I would guess (perhaps wrongly) that any "natural" inner product would satisfy at least some kind of continuity condition.

`$\Vert x\Vert^2$`

in the denominator, so that the supremum exists. But even with this correction, the supremum will ruin the linearity required for an inner product. The supremum of the sum of two functions is usually not the sum of their suprema. – Andreas Blass Mar 10 '12 at 1:53