Proving a determinant of a particular type is always null

It is true over any field, not just $\mathbb R$. Note that $\det\pmatrix{I&A\\ Q&B}=\det(B-QA)$. View $A$ and $B$ as two linear maps from a vector space $V$ to another vector space $W$ of the same dimension. View $Q$ as a linear operator on $W$. By changing the bases of $V$ and $W$ separately, we may assume that $A=I_r\oplus0$ where $r=\operatorname{rank}(A)$. Partition $B$ and $Q$ accordingly as $[B_1|B_2]$ and $[Q_1|Q_2]$, where $B_1$ and $Q_1$ each has $r$ columns. Then $B-QA=[B_1-Q_1|B_2]$ is singular for every $Q_1$. Hence $B_2$ has deficient column rank and so does $\pmatrix{A\\ B}=\pmatrix{\ast&0\\ \ast&B_2}$. Since the change of bases amounts to a transformation in the form of $\pmatrix{A\\ B}\mapsto\pmatrix{U&0\\ 0&U}\pmatrix{A\\ B}V$ for some invertible $U$ and $V$, the $\pmatrix{A\\ B}$ before change also has deficient column rank.

Denote $$M_Q = \begin{pmatrix} I_n & A \\ Q & B\end{pmatrix}, \quad \operatorname{col}(A,B) = \pmatrix{A\\B}. $$

---

The statement does hold with the additional assumption that $\ker(A) \subseteq \ker B$, i.e. the row-space of $A$ contains that of $B$.

Suppose for contradiction that $\operatorname{col}(A,B)$ has full rank. Let $U$ denote the column space of $A,B$. Let $P$ denote a matrix whose columns form a basis of $U^\perp$. By using column operations on $P$, we can bring $P$ to its column-echelon form, which is $$ P = \pmatrix{I_n\\ Q} $$ for some matrix $Q_*$. Because the columns of $P$ form a basis of $U^\perp$ and the columns of $\operatorname{col}(A,B)$ form a basis of $U$, we conclude that the columns of $M_{Q_*}$ form a basis of $\Bbb R^n$, which means that $M_{Q_*}$ is invertible and $\det(M_{Q_*}) \neq 0$.

Thus, $\operatorname{col}(A,B)$ indeed fails to have full rank if $\det(M_Q) = 0$ for all $Q$.