If ${\rm rank} \, X>1$, there exists a two-dimensional space $\mathcal{L}$ such that the restriction of $X$ to $\mathcal{L}$ is positive definite. The space of Hermitian operators, which vanish on $\mathcal{L}^\perp$, is 3-dimensional,thus, it contains a non-zero operator $Y$ for which ${\rm tr} \, Y={\rm tr} \, BY=0$. The operator $X+sY$ is positive semidefinite for real $s$ close to 0. We may suppose that ${\rm tr} \,AY\geqslant 0$, otherwise change $Y$ to $-Y$. Consider the maximal $s$ for which $X+sY$ is positive semidefinite (it exists, since $Y$ itself is not positive semidefinite because of ${\rm tr} \,Y=0$). Replace $X$ to $X+sY$. The value of ${\rm tr} \,AX$ increased (maybe not strictly), and the rank of $X$ strictly decreased. After finitely many such changes we get a matrix with rank at most 1.

The set over which you maximize is compact, thus the maximum is achieved, unless this set is empty. The latter happens if $\pm B$ is positive definite.

If ${\rm rank} \, X>1$, there exists a two-dimensional space $\mathcal{L}$ such that the restriction of $X$ to $\mathcal{L}$ is positive definite. The space of Hermitian operators, which vanish on $\mathcal{L}^\perp$, is 4-dimensional (3—dimensional in the real case), thus, it contains a non-zero operator $Y$ for which ${\rm tr} \, Y={\rm tr} \, BY=0$. The operator $X+sY$ is positive semidefinite for real $s$ close to 0. We may suppose that ${\rm tr} \,AY\geqslant 0$, otherwise change $Y$ to $-Y$. Consider the maximal $s$ for which $X+sY$ is positive semidefinite (it exists, since $Y$ itself is not positive semidefinite because of ${\rm tr} \,Y=0$). Replace $X$ to $X+sY$. The value of ${\rm tr} \,AX$ increased (maybe not strictly), and the rank of $X$ strictly decreased. After finitely many such changes we get a matrix with rank at most 1.

The set over which you maximize is compact, thus the maximum is achieved, unless this set is empty. The latter happens if and only if $\pm B$ is positive definite.