The idea is to apply a unitary congruence $$U=\dfrac{1}{\sqrt{2}}\begin{pmatrix}I&-I\\I&I\end{pmatrix}.$$ I consider here $\mathcal{B}$ to be hermitian, $\mathcal{B}=\mathcal{B}^*$, whereas the general case may be 'different'.
So $$R=UT(t)U^*=\begin{pmatrix}\mathcal{A}-\mathcal{B}&tI\\tI&\mathcal{A}+\mathcal{B}\end{pmatrix}.$$ Similarly $$ R-\lambda I=\begin{pmatrix}-\lambda I&(A-B)&tI&0\\(A^*-C)&-\lambda I&0&tI\\tI&0&-\lambda I &(A+B)\\0&tI&(A^*+C)&-\lambda I\end{pmatrix}. $$ Using the well known determinant formula for block matrices with a commuting off-diagonal block, you obtain that the eigenvalues $\lambda$ satifyy $$\det\left(\begin{pmatrix}-\lambda I&F\\G&-\lambda I\end{pmatrix}\begin{pmatrix}-\lambda I&2A-F\\2A^*-G&-\lambda I\end{pmatrix}-t^2I\right)=0.$$ Equivalently,= $$\det\begin{pmatrix}(\lambda^2-t^2)I +F(2A^*-G)&-2\lambda A\\-2\lambda A^*&(\lambda^2-t^2)I+G(2A-F)\end{pmatrix}=0.$$
The monotonicity of the eigenvalues follows and $-\lambda$ is also an eigenvalue since $$\begin{pmatrix}Q&X\\Y&Z\end{pmatrix}$$ is unitarily congruent to $$\begin{pmatrix}Q&-X\\-Y&Z\end{pmatrix}.$$
