In Gauss's law integral form, when to use Q_free, when to use Q_enclosed
You are correct.

Allow me to rigorously demonstrate what you have said using the following definitions.

1.)
$$\mathbf{D} = \epsilon \mathbf{E} = \epsilon_r \epsilon_0 \mathbf{E}$$
2.) Polarization

$$P = \epsilon_0 \chi_e \mathbf{E}\  = np_a[Cm^{-2}]$$
n is the average dipoles per \$m^3\$, \$p_a [Q m]\$ is the average dipole moment.

3.) Electric Susceptibility of a Material:
$$ \chi_e = \epsilon_r - 1$$
A constant of proportionality. It quantifies the material's ability to be polarized in an electric field. 

4.) Polarization Charge Density:

$$\oint \mathbf{P} \cdot d\mathbf{s} = - \int_v\rho_p dv = -q_p \Longleftrightarrow  \nabla \cdot \mathbf{P}= - \rho_p$$

Let \$q_f\$ be enclosed free charge. Let \$q_t\$ be enclosed total charge. Let \$q_p\$ be enclosed polarization charge.

$$q_f = \oint_s \mathbf{D} ds $$ and $$q_t = \oint_s \mathbf{\epsilon_0 E} ds $$

Then 

$$q_f = \oint_s \mathbf{D} ds = \epsilon_r \oint_s \mathbf{\epsilon_0 E} ds = (\frac{P}{\epsilon_0 \mathbf{E}} + 1 )  \oint_s \mathbf{\epsilon_0 E} ds  = \oint_s \mathbf{P} \cdot d\mathbf{s}  +  \oint_s \mathbf{\epsilon_0 E} ds = -q_p  + q_t$$