The magnetic part of the Lorentz force acts perpendicular to the charge's velocity, and consequently does zero work on it. Can we extrapolate this statement to say that such a nature of the force essentially makes its corresponding work independent of the choice of path, and hence that the magnetic force is conservative?
Not really, because the magnetic force is velocity dependent, not solely position dependent, so you can't extrapolate from knowing the integral along a path is zero to the conclusion that the force is the gradient of a potential.
What you can do is make an analog of the potential argument for the momentum components, so that the magnetic field is the curl of a vector potential. This argument can be made physically for conservation of momentum around a space-time loop, much like the conservation of energy follows from the integral of the force along a space-loop.
Well, the work is zero regardless of the path you take, so in this sense the magnetic force is (trivially) conservative.
Work done by the magnetic force is indeed zero.
$F_m=q[V \times B]$
$dW=q[V \times B].dr =q[V\times B]Vdt=0$
Consequently for work done along some arbitrary path from A to B$$=\int_A^B dW=0$$
Work done by the magnetic force is independent of path[and equal to zero for all such paths]
However we may have the following interesting consideration:
[We may choose $V_1$ and $V_2 $so that $V_1,V_2$ and $B$ are not in the same plane]
$=q[V_1 \times B]V_1+q[V_1 \times B]V_2+q[V_2 \times B]V_1+q[V_2 \times B]V_2$
$0=q[V_1 \times B]V_2+q[V_2 \times B]V_1$
The quantities $q[V_1 \times B]V_2 $and $q[V_2 \times B]V_1$ may not be zero individually[considering tha fact that$ V_1,V_2$ and $B$ do not lie in the same plane according to our choice] though their sum is zero.We might think of using either of them for some technological purpose.
So it is just a matter of definitions.