For your original question I'm pretty sure the answer is obviously no, because you can start with Popa's construction of a TLJ subfactor of $L(F_\infty)$ and then just change the factor in some boring way. For example, start with $N \subset M$ Popa's inclusion of $L(F_\infty)$ factors and then tensor it with the hyperfinite to get $N \otimes R \subset M \otimes R$. Unless I'm missing something, this won't change the standard invariant, and since the free group factors aren't McDuff this gives a different subfactor.

For the hyperfinite $\mathrm{II}_1$ basically everything about TLJ at index above 4 is open. In particular, it's open to determine for which indices it exists and it's open to determine for which indices it's unique. (Just to give a glimpse of how open things are, I think it's open whether there are any transcendental indices realized, whether there is any index at all that isn't realized, and whether there's any index which can be realized in infinitely many distinct ways.)

For the $L(F_\infty)$ case, my understanding was that it's not unique, though I don't myself know how to prove that. I don't know whether this has been proven or is just expected.