It is not true that $\tilde A$ maps trace class operators to trace class operators in general. For a counterexample, consider the maps $A:X\mapsto \mathrm{Tr}(X) e_{1,1}$. Then $\tilde A$ should send $e_{1,1}$ to $\mathrm{Id}_{\mathcal H}$, which is not trace class.

What is true in full generality is that, if $X$ is a Banach space and $A \in B(X)$, then its adjoint $A^*$ is a well-defined operator on the dual of $X$, and that the spectrum of $A$ and $A^*$ coincide.

In your setting, the dual of the space of trace class operators is $B(\mathcal H)$, so your operator $\tilde A$ extends to a bounded map on $B(\mathcal H)$, which has the same spectrum as $A$.

It is not true that $\tilde A$ maps trace class operators to trace class operators in general. For a counterexample, consider the maps $A:X\mapsto \mathrm{Tr}(X) \vert 1\rangle \langle 1 \vert$. Then $\tilde A$ should send $\vert 1\rangle \langle 1 \vert$ to $\mathrm{Id}_{\mathcal H}$, which is not trace class.

What is true in full generality is that, if $X$ is a Banach space and $A \in B(X)$, then its adjoint $A^*$ is a well-defined operator on the dual of $X$, and that the spectrum of $A$ and $A^*$ coincide.

In your setting, the dual of the space of trace class operators is $B(\mathcal H)$, so your operator $\tilde A$ extends to a bounded map on $B(\mathcal H)$, which has the same spectrum as $A$.