Given a system of $N$ point-particles with positions
${\bf r}_1, \ldots , {\bf r}_N$; with corresponding virtual displacements $\delta{\bf r}_1$, $\ldots $, $\delta{\bf r}_N$; with momenta ${\bf p}_1, \ldots , {\bf p}_N$; and with applied forces ${\bf F}_1^{(a)}, \ldots , {\bf F}_N^{(a)}$. Then D'Alembert's principle states that
$$\tag{1} \sum_{j=1}^N ( {\bf F}_j^{(a)} - \dot{\bf p}_j ) \cdot \delta {\bf r}_j~=~0. $$
The total force
$${\bf F}_j ~=~ {\bf F}_j^{(a)} +{\bf F}^{(ec)}_j+{\bf F}^{(ic)}_j
+ {\bf F}^{(i)}_j + {\bf F}_j^{(o)}$$
on the $j$'th particle can be divided into five types:
applied forces ${\bf F}_j^{(a)}$ (that we keep track of and that are not constraint forces).
an external constraint force ${\bf F}^{(ec)}_j$ from the environment.
an internal constraint force ${\bf F}^{(ic)}_j$ from the $N-1$ other particles.
an internal force ${\bf F}^{(i)}_j$ (that is not an applied or a constraint force of type 1 or 3, respectively) from the $N-1$ other particles.
Other forces ${\bf F}_j^{(o)}$ not already included in type 1, 2, 3 and 4.
Because of Newton's 2nd law ${\bf F}_j= \dot{\bf p}_j$, D'Alembert's principle (1) is equivalent to$^1$
$$\tag{2} \sum_{j=1}^N ( {\bf F}^{(ec)}_j+{\bf F}^{(ic)}_j+{\bf F}^{(i)}_j+{\bf F}_j^{(o)}) \cdot \delta {\bf r}_j~=~0. $$
So OP's question can essentially be rephrased as
Are there examples in classical mechanics where eq. (2) fails?
Eq. (2) could trivially fail, if we have forces ${\bf F}_j^{(o)}$ of type 5, e.g. sliding friction, that we (for some reason) don't count as applied forces of type 1.
However, OP asks specifically about internal forces.
For a rigid body, to exclude pairwise contributions of type 3, one needs the strong Newton's 3rd law, cf. this Phys.SE answer. So if these forces fail to be collinear, this could lead to violation of eq. (2).
For internal forces of type 4, there is in general no reason that they should respect eq. (2).
Example: Consider a system of two point-masses connected by an ideal spring. This system has no constraints, so there are no restrictions to the class of virtual displacements. It is easy to violate eq. (2) if we count the spring force as a type 4 force.
Reference:
H. Goldstein, Classical Mechanics, Chapter 1.
--
$^1$It is tempting to call eq. (2) the Principle of virtual work, but strictly speaking, the principle of virtual work is just D'Alembert's principle (1) for a static system.
Actually, at least for a single point subjected to a friction force $F= -\gamma v$ and other forces associated with a potential $U(t,x)$ there exists a Lagrangian: $${\cal L}=e^{t\gamma/m}\left(\frac{m}{2}\dot{x}^2 -U(t,x)\right)\:.$$
The point is that this Lagrangian is not of the form $T-U$, nevertheless it gives rise to the correct equation of motion, the same obtained by using the Rayleigh dissipation function you mentioned.
This Lagrangian however cannot be produced by direct application of the principle of virtual works you mention.
Best Answer
1) According to usual terminology we wouldn't call a sliding friction force a constraint force as it doesn't enforce any constraint. (No pun intended.) In other words, a sliding friction does not by itself constrain the particles to some constraint subsurface, i.e., the particles can still move around everywhere.
On the other hand, rolling friction and static friction may actually impose a constraint, so they can be constraint forces.
2) In more detail, Goldstein says on p.17 in Chapter 1 of the book Classical Mechanics that
Goldstein goes on to say that
What he has in mind is, that we can still at least analyze and study many systems of fundamental/microscopic point particles (which is anyway the most important case!) with the principle of virtual work, because there are often no sliding friction forces involved down at these scales.
In particular, Goldstein does not imply that sliding friction forces are not important in macroscopic systems.
3) Later in chapter 1, Goldstein confronts us with another problem with sliding friction forces. They cannot be described with the help of a velocity-dependent potential $U$ but only in terms of Rayleigh's dissipation function ${\cal F}$. This is related to the fact that there is no action principle for systems with sliding friction forces.