Let't consider 2 point particle graviting the one around the other. Can that gravitational field be considered conservative? I can go from A to B and then, after a time $\Delta t$ come back to A with a non zero work. I imagine the objection: the closed loop has to be done at infinite speed. But if this is the case why in "Why can't conservative forces depend on velocity?" It is said that conservative forces are time independent? I'm confused. Can the time dependent force field I metioned be considerated conservative? Isn't its potential $-G\frac{m_1 m_2}{d_{12}}$ even if both masses are finite and so both moving?
Do not confuse an external force with an internal force. You can consider both particles to be part of the dynamical system, but then there is no external force. The Hamiltonian is:
$$
H = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} - \frac{Gm_1m_2}{|\vec r_1 - \vec r_2|}\;.
$$
The force on particle 1 is:
$$
-\frac{\partial H}{\partial \vec r_1} = \frac{Gm_1m_2(\vec r_2 - \vec r_1)}{|\vec r_1 - \vec r_2|^3}\;.
$$
The force on particle 2 is:
$$
-\frac{\partial H}{\partial \vec r_2} = \frac{Gm_1m_2(\vec r_1 - \vec r_2)}{|\vec r_1 - \vec r_2|^3}\;,
$$
(as expected from Newton's third law).
This Hamiltonian is a conserved quantity when the system is closed (both particles 1 and 2 Hamiltonian-dynamical):
$$
\frac{dH}{dt} = 0
$$
This conservation follows directly from the equations of motion:
$$
\frac{\partial H}{\partial \vec r_1} = -\dot {\vec p_1}\;,
$$
$$
\frac{\partial H}{\partial \vec p_1} = \dot {\vec r_1}\;,
$$
etc for particle 2.
You can see this explicitly because:
$$
\frac{dH}{dt} =
\frac{\partial H}{\partial \vec r_1}\cdot \dot{\vec r_1}
+\frac{\partial H}{\partial \vec r_2}\cdot \dot{\vec r_2}
+\frac{\partial H}{\partial \vec p_1}\cdot \dot{\vec p_1}
+\frac{\partial H}{\partial \vec p_2}\cdot \dot{\vec p_2}
$$
$$
=\frac{\partial H}{\partial \vec r_1}\cdot \frac{\partial H}{\partial \vec p_1}
+\frac{\partial H}{\partial \vec r_2}\cdot \frac{\partial H}{\partial \vec p_2}
-\frac{\partial H}{\partial \vec p_1}\cdot \frac{\partial H}{\partial \vec r_1}
-\frac{\partial H}{\partial \vec p_2}\cdot \frac{\partial H}{\partial \vec r_2}
$$
$$
=0
$$
Update based on OP's update
In other words. Lets consider a situation like the following: consider a test particle in the gravitational field of a moving object. If we move the particle along a loop starting and ending at A, then the gravitational potential at A has changed by the time the particle comes back, and so the net work done on the particle is non-zero. So this gravitational field is not conservative? I must say that gravitational field is conservative only if it is time independent? Is it in some way meaningfull to speak about potential energy of a time dependent field?
If you treat one of the masses as external to the system and move it around in a non-trivial time-dependent way, it will look like a time-dependent potential to the other mass. Then you will be treating that other mass as a single particle in a time-dependent field.
Suppose we only wanted to treat one particle dynamically, and the other as a parameter that we move around with arbitrary time dependence. Then we would have:
$$
H = \frac{p}{2m} - \frac{GmM}{|\vec r - \vec R(t)|}\;,
$$
where I put $r_1 \to r$, $r_2 \to R$, etc.
In either case (whether particle 2 is dynamical or not) we could say the force on particle 1 is conservative since either:
$$
\vec \nabla_1 \times \vec F_1 = 0\;,
$$
or
$$
\vec \nabla \times \vec F = 0\;.
$$
But the energy is not conserved in the latter case in general, since:
$$
\frac{dH}{dt} = \frac{\partial H}{\partial \vec R}\cdot \dot{\vec R}\;.
$$
We have to supply or remove energy to keep the non-dynamical particle 2 moving about in its fixed time-dependent path $\vec R(t)$.
