To calculate this expectation it is helpful to start from the polar decomposition
$$Q= \left(\begin{array}{cc}U'&0\\ 0&V'\end{array}\right) \left(\begin{array}{cc}\sqrt{1-T}&\sqrt{T}\\ \sqrt{T}&-\sqrt{1-T} \end{array}\right) \left(\begin{array}{cc}U&0\\ 0&V\end{array}\right) $$ where $U,V,U',V'$ are four unitary matrices and $$ T=\left(\begin{array}{cc}T_1&0\\ 0&T_2\end{array}\right)$$ is a diagonal matrix with diagonal elements $0\leq T_n\leq 1$.
You seek the expectation value of the matrix $$M=[I-U^{\rm H}(1-T)U]^{-1}=U^{\rm H}T^{-1}U$$ (I have used that $U^{\rm H}U=I$.)
The Haar distribution for $Q$ implies a Haar distribution for $U$, and moreover implies for $T_1,T_2$ the following distribution [*]
$$P(T_1,T_2)=6(T_1-T_2)^2,\;\;0\leq T_n\leq 1$$
The marginal distribution for $T_1$ is
$$P(T_1)=2-6T_1+6T_1^2$$
So you see that the expectation value of $M$ diverges: $E({\rm Tr}\,M)=2E(1/T_1)=\infty$.
[*] See Equation (2.9) of this review.