Your gap is too wide. Make it very, very very narrow. And rolled into a cylinder. Like a real capacitor.
Yes, +q could be less than -q, but only if the attraction/repulsion effects of electrons in the connecting wires were nearly as large as the attraction/repulsion down between the capacitor plates. (In that case the plates wouldn't be a near-perfect electrical shield for the fields produced by the wires.) But with real-world capacitors, this doesn't happen, and instead the field between the plate is totally enormous compared to the tiny fields produced by electrons in the wires. If +q only differs from -q by a millionth of a percent, we ignore it. See Engineer's capacitor vs. Physicist's capacitor, a split metal ball, versus two separate balls.
For capacitors used in circuitry, if we dump some charge on one capacitor terminal, exactly half of it will seemingly migrate to the other terminal. Weird. But "physicist-style capacitors" with small, wide-spaced plates are different, and an extra electron on the wire will make +q not equal to -q.
In detail: if the capacitance across the plates is 10,000pF, and the capacitance to Earth of each wire and plate is 0.01 pF, then the opposite plate's charges will ignore any small +q and/or -q on the connecting wires. The attraction/repulsion of electrons in the wires doesn't significantly alter the enormous +q and -q on the inner side of the capacitor plates.
Engineers use real-world components: wide capacitor plates with very narrow gaps; gaps the thickness of insulating film. But if you were a physicist, your capacitors might be metal spheres with large gaps between, or metal disks where the space between the plates was large when compared to their diameter. (Or you'd draw a capacitor symbol where the gap between plates was enormous and easy to see.) In this case the attraction/repulsion of electrons on the connecting wires would have an effect on the balance of +q -q between capacitor plates.
PS
Another weird concept: make a solid stack of thousands of disc capacitors: foil disk, dielectric disk, foil disk, etc. Use half-inch wide disks, and stack them up into a narrow foot-long rod. Now connect one end to 1,000 volts. The same kilovolt will appear on the other end! The rod is acting like a conductor. Yet its DC resistance is just about infinite. Series capacitors! Each little capacitor induces charge on the next and the next, all the way to the end.
Gauss Law. Polarization Vector
The Gauss Law brings the local relation between the electric field and the sources.
The main sources of electric field are free charges, but we can also consider the contribution of the field produced by the polarized material.
$$
\nabla\cdot\mathbf{E}(\mathbf{r}) =\dfrac{\rho_l(\mathbf{r}) + \rho_P(\mathbf{r})}{\varepsilon_0}
$$
where \$\rho_l\$ is the free charge contribution, and \$\rho_P\$ is the contribution due to polarization. But
$$
\rho_P(\mathbf{r})=-\nabla\cdot\mathbf{P}(\mathbf{r})
$$
where \$\mathbf{P}(\mathbf{r})\$ is the Polarization Vector. Then
$$
\nabla\cdot\mathbf{E}(\mathbf{r})=\dfrac{\rho_l(\mathbf{r})-\nabla\cdot\mathbf{P}(\mathbf{r})}{\varepsilon_0}
$$
The free charge density is
$$
\rho_l(\mathbf{r})=\nabla\cdot\left(\varepsilon_0\,\mathbf{E}(\mathbf{r})+\mathbf{P}(\mathbf{r})\right)
$$
remember that \$\nabla\cdot\mathbf{D}(\mathbf{r})=\rho_l(\mathbf{r})\$, we can write the general form of the Gauss Law:
$$
\mathbf{D}(\mathbf{r})=\varepsilon_0\,\mathbf{E}(\mathbf{r})+\mathbf{P}(\mathbf{r})
$$
The displacement vector \$\mathbf{D}(\mathbf{r})\$ is the combination of the applied field \$\mathbf{E}\$ and induced field \$\mathbf{P}\$ in the material by the polarization of its molecules. The polarization of a material depends on the external field, and in turn creates an induced field which overlaps the external field. Then there is a relationship between these fields, in particular between the polarization vector and total field (the field that can be measured).
For linear dielectrics (which are the most technological interest materials) applies: \$\mathbf{P}(\mathbf{r})=\chi_e\,\varepsilon_0\,\mathbf{E}(\mathbf{r})\$ and then
$$
\mathbf{D}(\mathbf{r})=\varepsilon_0(1+\chi)\mathbf{E}(\mathbf{r})=\varepsilon_0\,\varepsilon_r\,\mathbf{E(r)}=\varepsilon\,\mathbf{E(r)}
$$
where \$\chi\$ is the dielectric susceptibility of the material. \$\varepsilon=\varepsilon_0\,\varepsilon_r=\varepsilon_0(1+\chi)\$ is the permittivity of material, and \$\varepsilon_r\$ is the relative permittivity.
The greater the permittivity of the material, is polarized more strongly and electrical effects are greater.
Best Answer
The flux density is "created" by the charged particles just as our friend Maxwell tells us: \$\nabla \cdot \vec{D} = \rho \$. The other way to look at this relationship is to think that the net charge contained in any volume can be calculated by taking a surface integral of \$\vec{D}\$ over the surface of this volume. Note that \$\vec{D}\$ is a vector quantity.
See the three steps in the drawing below, red marks positive charge, blue negative, and black is the flux density \$\vec{D}\$:
For a point charge, the \$\vec{D}\$ can be calculated just by drawing a sphere around the charge and saying that at distance \$r\$ the surface density is \$ D = Q / 4\pi r^2\$.
For a single (very large) plate that has a charge density of \$ 7~C/m^2\$, the \$\vec{D}\$ can be calculated by taking a small (square meter) section of the plate. We can see that the charge contained in the section is 7 C and the surface area over which the 2 square meters (one surface on both sides of the plate). So \$D = \frac{7~C/m^2}{2} = 3.5~C/m^2\$.
In phase three, we take a second plate that has a charge of \$-7~C/m^2\$. This second plate creates a \$\vec{D}\$ around itself just like the first one, just flux going in the opposite direction, that is, towards the plate. The vector sum of these two fields is \$\vec{D} = 7~C/m^2\$ towards the negatively charged plate. The sign of the answer is just a matter of how you set your coordinate system. Normally we say that z-axis points upwards and in your question, the top plate is the positively charged one. As a conclusion, the \$\vec{D} = - 7 C/m^2 \vec{\hat{z}}\$, where \$\vec{\hat{z}}\$ is the unit vector.