Does d'Alembert principle hold for non-conservative forces?

All D'Alembert's Principle requires is that displacements under the force in question are reversible, so there are many non-conservative forces that exhibit this property.

Take, for example, the force $\mathbf{F}(x,y)=(y,-x)$. This force is not conservative, as it has nonzero curl at the origin and therefore cannot be a gradient of a scalar function. However, unlike friction, the direction of this force does not depend on the direction of travel, so any work done along any path can be reversed by traveling in reverse along the same path.

For an easy demonstration, consider paths consisting of some segment of a circle around the origin of radius $R$. The path has a length $s = R\Theta$, where $\Theta$ is the angle the path sweeps out. If we travel clockwise on this path, since $\mathbf{F}$ is always parallel to $\mathbf{ds}$, and has constant magnitude $\sqrt{x^2+y^2}=R$, the work done is simply

$$W_1 = |F|R\Theta=R^2\Theta$$

If we travel in reverse (counterclockwise) on this path, $\mathbf{F}$ is now pointing in the opposite direction to $\mathbf{ds}$, so the work done in getting back to our start point is

$$W_2 = -|F|R\Theta = -R^2\Theta$$

As you can see, $W_1+W_2$ vanishes for any $\Theta$ and $R$, so the work along any such circular path can be reversed. It's relatively intuitive at this point to extend this demonstration to encompass all paths.

1. Firstly, it depends on whether you put the non-conservative force in the bin of applied forces or not, cf. my Phys.SE answer here. To make your question non-trivial, let us assume that you do not put the non-conservative force in the bin of applied forces.

2. Secondly, it depends on your definition of conservative force. If we follow the definition of Wikipedia (July 2017), the magnetic Lorentz force $$\vec{F}~=~\vec{v}\times \vec{B}\tag{1}$$ is not a conservative force, since it depends on the velocity $\vec{v}$. On the other hand, it produces no work, so (1) cannot violate d'Alembert's principle. For this (and other) reasons I propose in my Phys.SE answer here to relax the conventional definition of conservative forces to allow some velocity-dependent forces.