Q1: No. You gave a proof above: $\overline{\langle f,\bar g\rangle} = \operatorname{Tr}(f\otimes g)$.
Q2: No, if $\operatorname{Tr}$ is supposed to be continuous. Namely, $\mathcal S(\mathbb R^k)\subset L^2(\mathbb R^k)$ continuosly, and $L^2(\mathbb R^k)\subset \mathcal S'(\mathbb R^k)$ countinuous and dense. So any continuous trace would immediately lead to a contradiction to Q1.
