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.
Each capacitor stores energy which is conserved. The energy stored in one of the capacitors is
$$\frac{1}{2}CV^2 = \frac{1}{2}(2\:\mathrm F) (2\:\mathrm V)^2 = 4\:\mathrm J$$
for a total of \$8\:\mathrm J\$ of stored energy. Whether the capacitors are placed in parallel or series, the amount of energy stored is the same.
If the charged capacitors are placed in parallel appropriately, the voltage across the combination is \$2\:\mathrm V\$ and the energy stored is \$8\:\mathrm J\$.
Thus the equivalent capacitance is
$$C_\text{EQ} = 2\frac{8\:\mathrm J}{(2\:\mathrm V)^2} = 4\:\mathrm F$$
If the charged capacitors are placed in series appropriately, the voltage across the combination is 4V and the energy store is \$8\:\mathrm J\$.
Thus, the equivalent capacitance is
$$C_\text{EQ} = 2\frac{8\:\mathrm J}{(4\:\mathrm V)^2} = 1\:\mathrm F$$
Yes, there is a \$3\:\mathrm F\$ difference but asking "what happened to the other 3 farads?" is like asking "what happened to the other 3 ohms?" when comparing series and parallel connected 2 ohm resistors.
No capacitance has 'vanished'. Both capacitors still have \$2\:\mathrm F\$ each of capacitance. What has changed is the configuration of the capacitors and, thus, the equivalent capacitance as seen by an external circuit.
Best Answer
At the current state of our universe, charge is conserved. (This wasn't necessarily always the case. See this article on dark matter, for example, discussing the possibility that charged particles created shortly after the Big Bang lost their electric charge during the inflationary period.) This means that the sum of two relative charges held by the two capacitors before being connected to each other must be the same as the relative charge of the combined capacitor after being connected. When you place two capacitors in parallel, the total charge of the final system is the sum of the two original charges on the two earlier systems. In short, \$q_{_{tot}}=q_{_1}+q_{_2}\$.
Since \$q_{_{1}}=C_{_{1}}\cdot V_{_{1}}\$ and \$q_{_{2}}=C_{_{2}}\cdot V_{_{2}}\$ and also that \$q_{_{final}}=C_{_{final}}\cdot V_{_{final}}\$; and since we also know that when two capacitors are placed in parallel that the total system capacitance is the sum of the two original system capacitances, or \$C_{_{final}}=C_{_{1}}+ C_{_{2}}\$; it then follows directly that:
$$q_{_{final}}=q_{_1}+q_{_2}=C_{_{1}}\cdot V_{_{1}}+C_{_{2}}\cdot V_{_{2}}=\left(C_{_{1}}+ C_{_{2}}\right)\cdot V_{_{final}}$$
It's easy then to re-arrange this so that:
$$V_{_{final}}=\frac{C_{_{1}}\cdot V_{_{1}}+C_{_{2}}\cdot V_{_{2}}}{C_{_{1}}+ C_{_{2}}}$$
Note that the energy of the final system is not the sum of the energy of the original two systems:
$$\frac12 C_{_{final}}\,V_{_{final}}^{^2}\ne \frac12 C_{_{1}}\,V_{_{1}}^{^2}+\frac12 C_{_{2}}\,V_{_{2}}^{^2}$$
In fact, the total energy will be smaller by this factor:
$$\Delta W = W_{_{final}}-\left(W_{_1}+W_{_2}\right)=-\frac12 \left(V_{_1}-V_{_2}\right)^2\frac{C_{_{1}}\cdot C_{_{2}}}{C_{_{1}}+C_{_{2}}}$$
This is true regardless of how that difference is expended. You may want to read this hyperphysics page on where the energy loss goes when charging up a capacitor. This lost energy can be through realistic resistance dissipation as heat. But as you assume a smaller and smaller connection resistance, so that the current is very much larger and the transfer time much shorter, then an increasing amount of the lost energy goes into electromagnetic radiation.