Is mechanical energy conserved when charges accelerate? Suppose we have two identical charges $q_1$ and $q_2$ and there is some distance $d$ between them. Imagine they are not allowed to move at first, but suddenly "we let them go". Then, they will start to move away from each other with a non-constant acceleration, due to the time-varying force acting on them. Namely, Coulomb's law tells us that each one experiences a force such that its magnitude is: $$F(t)=\frac{q_1q_2}{4\pi \epsilon_0}\frac{1}{d^2(t)}$$
What can we do to find the speed $v(t)$ at which the charges will be moving at any instant $t$? Well, at first I thought of using the conservation of mechanical energy. The electric potential energy of the system at instant $t$ would be $$U_e(t) = \frac{q_1q_2}{4\pi \epsilon_0 d(t)}$$
And therefore the speed could be found by calculating the amount of that potential energy that is transformed to kinetic energy to conserve the mechanical energy of the system.
However, this is where my confusion arose. The charges are accelerating, and therefore there is radiation going on. This means that there is a time-varying electric field that generates a time-varying magnetic filed, which generates a time-varying electric field, and so on. These contributions to the electric field are not conservative, as they curl is not zero. This would mean that thinking of electric potential energy makes no sense, as the electric field in this case would not be electrostatic and, therefore, would not be conservative (leading to the idea of an "electric potential" being nonsense).
So what's going on? Is mechanical energy conserved? If yes, how can it be possible given that the fields are non-conservative?  
 A: First calculate the acceleration of the particles due to the electrostatic effect, assuming the radiation effect is negligible. The method you outlined is appropriate. Or just calculate the force on them from the electric field and divide by their mass.
Then calculate how much radiation this would produce, assuming it doesn't change their acceleration significantly.
Now compare that radiation energy to the electrostatic potential energy and kinetic energy terms from your initial solution. 
If it's negligible, you're done. The electrostatic solution is close enough. (The definition of "negligible" and "close enough" may depend on how you intend to use the results of this calculation).
If the radiation energy is not negligible, then you're likely going to need to use a numerical simulation of some kind to obtain a closer approximation to the true answer.
A: Mechanical energy is not conserved, this system has electromagnetic interaction so part of energy will be electromagnetic.

The charges are accelerating, and therefore there is radiation going on. This means that there is a time-varying electric field that generates a time-varying magnetic filed, which generates a time-varying electric field, and so on.

Both electric and magnetic field are time-dependent, but this does not mean that one creates the other. In the simplest variant of this setup (fields are given by retarded solution), both fields are (different) functions of past motion of the particles.

These contributions to the electric field are not conservative, as
  they curl is not zero. This would mean that thinking of electric
  potential energy makes no sense, as the electric field in this case
  would not be electrostatic and, therefore, would not be conservative
  (leading to the idea of an "electric potential" being nonsense).

Hold on. It is true that total electric field is not conservative. But the concept of electric potential energy is still valid, because the electric field has easy-to-determine Coulomb part. Coulomb energy, in general situation, does not give exactly total EM energy, but the concept can be used anyway. There are just other contributions, like magnetic energy, and the other part of electric energy that is not in Coulomb energy.

So what's going on? Is mechanical energy conserved? If yes, how can it be possible given that the fields are non-conservative? 

Initial EM energy, given by the Coulomb formula, is continuously being transformed into kinetic energy of particles and other non-Coulombic EM energy. Part of this EM energy goes unavoidably away to the environment (waves to infinity).
A: 
Namely, Coulomb's law tells us that each one experiences a force such that its magnitude is: $$F(t)=\frac{q_1q_2}{4\pi \epsilon_0}\frac{1}{d^2(t)}$$

This is exactly wrong! :) As you'd recall, Coulomb's law is a law applicable to an electrostatic situation. When the charges are allowed to move, we have a patently dynamical situation and one cannot apply Coulomb's law. One would have to use the full Maxwell equations to solve for the electric and magnetic fields produced by one charge at the position of the other charge and then apply Lorentz force law to ultimately find the force experienced by each of the charges.
One can find the electric and magnetic fields produced by a generically moving charge particle using, for example, the Liénard–Wiechert potential. However, you'd get a pretty coupled system of differential equations given that each of the charges is moving and would be producing a force on the other particle given by the Liénard–Wiechert potential which requires the velocity and position of the source at a retired time. I'd assume you will have to solve it numerically. 

The charges are accelerating, and therefore there is radiation going on.

This is precisely correct. The total energy of the system would obviously be conserved but a part of that energy would be in the form of radiation which travels off to infinity and thus cannot be thought of as contributing to the potential energy between the two particles. So, yes, one cannot use the conservation of mechanical energy in this situation, however, energy conservation still applies. In other words, $$\int dV \bigg( \frac{1}{2}\epsilon_0 E^2+\frac{1}{2\mu_0}B^2\bigg)+\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$
would still be conserved (assuming non-relativistic speed for particles). Just that a part of that energy in the field will be in the form of radiation. You won't be really able to use this effectively for you'd need to solve for $E$ and $B$ using the full dynamical machinery of Maxwell's equations to actually compute the integral. 
Finally, as @ThePhoton has pointed out, you can do an approximate calculation so long as your estimation of how much energy would be lost in radiation is low compared to the total mechanical energy in the initial state. 
