Yes, at least if $Q(x,y)$ is locally integrable in $y$.
Given two integral kernels $Q(x,y)$ and $Q'(x,y)$ that agree on a neighborhood of the diagonal in $X \times X$, then if one is a parametrix, so is the other one. Hence, you can always restrict a parametrix to have support in an arbitrary neighborhood of the diagonal. Just multiply any given parametrix $Q(x,y)$ by a smooth function $\chi(x,y)$ that has sufficiently small support but is still identically $1$ on a smaller neighborhood of the diagonal.
Since $Q(x,y)$ is locally integrable, you can choose the neighborhood small enough so that $\int_X |Q(x,y)| dx < C$ and $\int_X |Q(x,y)| dy < C$. Then, by the Schur test, $Q$ is bounded on $L^2(X)$.
Each of the operators $T$ and $PT$ is also represented by a locally integrable integral kernel. Thus, you can make the exact same argument for each of them, possibly restricting the support of $Q$ even closer to the diagonal, so that $T$ and $PT$ both satisfy the Schur test.
Finally, by your definition $PQ = 1+T$, hence it is automatically bounded.
If $Q(x,y)$ is not locally integrable, the above argument may not work, since there might not exist a sufficiently small neighborhood of the diagonal that will make all the $y$-integrals bounded.
Update: However, take a look at Thm VI.2.1 of Stein's Harmonic Analysis (PUP, 1993). It seems to directly address the $L^2$ boundedness of (scalar) pseudodifferential operators of a certain class, that might cover your operators.