Will the two capacitors be charged in this circuit? Is such a circuit possible to exist with the two capacitors charging?
Because if we considered the outer loop using Kirchhoff's law we get:
ℰ = q1/C1 + q2/C2
But since the two capacitors are initially uncharged, the two terms of the equation increase, that is, q1 and q2 increase, while they equal ℰ which is a constant.
How's that possible? and will the capacitors be charged? also how could we calculate the time constant for it?

 A: As drawn, the circuit, assuming ideal circuit elements, is problematic for the reason you've deduced (KVL gives a contradiction).  One interpretation is that there is infinite large current for an infinitesimal time which instantaneously charges the capacitors to their final steady state voltages.
To gain some insight, add a resistance $r$ in series with the battery; this models the internal resistance of a physical battery.  You will find that the initial battery current is equal to $\mathcal{E}/r$ and that it decays to a steady state value of $\mathcal{E}/(R_1 + R_2 + r)$.
Thus, see that as $r \rightarrow 0$, the initial current goes to infinity but this is clearly unphysical, no physical voltage source can supply arbitrarily large current. 
In fact, there are other mechanisms such as the inescapable inductance of the loop and radiation resistance that must be included in the model in the case that $r$ is 'small enough'.
In summary, it is possible (and well known in the EE community) that one can draw circuit diagrams that, assuming ideal circuit elements, lead to contradictions, e.g., two different parallel connected voltage sources.  The key is to understand that, in order to model physical circuits, one must often insert additional ideal circuit elements such as, in this case, a resistor in series with the battery to model the finite short circuit current capability.
A: No, that circuit cannot exist in that regime. You are neglecting the internal resistance of the wires between the voltage source and the capacitors, and if the capacitors are discharged (in which case the voltage over them is zero) that's no longer a good approximation. You therefore need to insert a small resistance on either side of the voltage source, which will then govern the $RC$ charging time of the capacitors.
