Let R be a commutative ring, and $A$ an $R$-algebra (possibly non-commutative). Then $A$ is separable if it is finitely generated (f.g.) projective as an $(A \otimes_R A^{\mathrm{op}})$-algebra. Suppose further that $A$ is f. g. projective as an $R$-module. Does this imply that $A$ is a (symmetric) Frobenius algebra?

There are lots of equivalent definitions of a Frobenius algebra. One (assuming $A$ is a f.g. projective R-module) is that there exists an $R$-linear map $\mathrm{tr}: A\to R$, such that $b(x,y) := \mathrm{tr}(ab)$ is a non-degenerate.

I know that the answer is yes when $R$ is a field. What about other rings?

I am not an expert on algebras, but this question is related to understanding obstructions for extended TQFTs, and so I am very interested in knowing anything I can about it.