If $e^{itA}$ is a special unitary matrix, what can I say about A? I know that if $e^{itA}$ is unitary, A must be skew-hermitian. I also know that $A$ must be traceless, since the determinant of $e^{itA}$ is $1$. What can I say about the determinant of A?

Presumably $t$ is a nonzero real number. Since matrix exponentials of non-trivial Jordan blocks are not diagonalisable over $\mathbb C$, if $e^{itA}$ is unitary, $A$ must be diagonalisable and every eigenvalue $\lambda$ of $A$ satisfies $|e^{it\lambda}|=1$. Hence each $\lambda$ is real, i.e. $A$ is a diagonalisable matrix with real eigenvalues, and $|\det(e^{itA})|=1$ because $e^{itA}$ is unitary.

However, we cannot infer that $A$ is traceless or Hermitian. E.g. for any two integers $m$ and $n$ with $|m|\
e|n|$, the matrix $$ A=\pmatrix{2m\pi&1\\\ 0&2n\pi} $$ is neither traceless nor symmetric, but $e^{iA}=I$ is real orthogonal. The determinant of $A$ is also unbounded in this example.