Assuming the system and probe are initially uncorrelated, the initial density matrix is
$$ \rho(0) = \rho_S(0) \otimes \rho_P(0). $$

After interaction for a time $t$, the system and probe are entangled
$$ \rho(t) = e^{-i H_{\text{full}}t}\rho(0)e^{iH_{\text{full}}t} \equiv \rho_{SP}.$$

Then the observable $\hat Q$ is measured, which can be written
$$ \hat Q = \sum_q q\hat\Pi_q,\qquad\text{with}\quad\hat\Pi_q=\sum_j|{q,j}\rangle\langle{q,j}|,$$
i.e., $\hat Q$ has a discrete spectrum (continuous involves replacing the sum with an integral), where the eigenvalue $q$ has an eigenspace spanned by $|q,j\rangle$ (it is degenerate if there is more than on $j$, which is the typical situation).

In a *projective* measurement of $\hat Q$, you obtain the eigenvalue $q$ with a probability $\text{Tr}[\rho(t)\hat Q]$.
The state of your system is then
$$\rho_q=\frac{\hat\Pi_q\rho(t)\hat\Pi_q}{\text{Tr}[\rho(t)\hat\Pi_q]}. $$

If you're not sure what you have measured (what Norbert Schuch called "classical scrambling" in his comment), then your state is
$$ \rho_{q_{\text{readout}}} = \int dq \, P(q_{\text{readout}}|q)\,\rho_q. $$
(or a sum if your probability distribution is discrete.)
In this case, if you ignore your measurement record, the state of your system is 
$$\rho_{\text{ignore}} = \int dq\,\rho_q.$$
This is the *unconditional state*.
In this case (and in the former) all coherences between different $q$ subsectors are gone (if you write the density matrix in a basis where different blocks correspond to different $q$, then all off-diagonal blocks are zero).

Finally, you could consider the case in which the measurement is weak. This is the case in which we really should know more about the measurement. Assuming everything is continuous, you can write down a family of *Kraus operators*
$$\hat{\Upsilon}_{q_r}=P(q_{\text{r}}|\hat Q), $$
where you just replaced the number $q$ with the operator $\hat Q$ in your expression above. In this case the state of your system post-measurement is
$$ \rho_{q_r} = \frac{\hat\Upsilon_{q_r}\rho(t)\hat\Upsilon_{q_r}}{\text{Tr}[\rho(t)\hat\Upsilon_{q_r}^\dagger\Upsilon_{q_r}]}. $$
When you now look at the unconditional state by integrating over $q_r$, you'll find it's not block diagonal!