The claim is false as stated. Let $n = 2$, consider $X_n = \begin{pmatrix} -n & 0 \\ 0 & -n \end{pmatrix}$. Since $\det X_n = n^2 > 0$ we have that $X_n \in D$. By positive homogeneity of norms $\|X_n\| = n \|X_1\| \nearrow \infty$. But $\ln \det X_n = 2 \ln n$, while $\operatorname{tr} X_n = -2n$, so $F(X_n) = - 2n - 2 \ln n \searrow -\infty$.
Edit [A slightly different counterexample]:
Again take $n = 2$. Take $X_n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$. We have that the trace of $X_n$ is always 2, while the determinant is always 1. So $F(X_n) = 2$ for all $n$. Both the Frobenius norm and the max norm on $X_n$ satisfy $\|X_n \| \geq n$ and this gives another family of counterexamples.
Furthermore, this example gives also an illustration of why you need to be precise about $X_k \to \partial D$. Observe that the matrix $Y_n = \begin{pmatrix} 1 & n \\ 1/n & 1\end{pmatrix}$ has vanishing determinant, and hence belongs in $\partial D$. In any norm $ \|X_n - Y_n\| = \left\| \begin{pmatrix} 0 & 0 \\ 1/n & 0\end{pmatrix} \right\| \searrow 0 $ so we in fact have not only do $X_n$ blow up in norm, it also asymptotically approaches $\partial D$!
(On the other hand, the statement that if $X_n \to X_0\in \partial D$ we have $F(X_n) \nearrow \infty$ is true by virtue of continuity of the trace function on $M_n$.)