I don't know the exact answer, but my guess is that it is approximately true for JWST for two reasons:
For small perturbations from L2, there are linear restoring forces that result in simple harmonic motion. This is true along the axis of rotation (z-axis) and also in the direction of rotation (y-axis). In the ecliptic plane (x-y plane), the Coriolis force acts to automatically bend the resulting simple harmonic motion into an ellipse.
So, at least for small perturbations, there is an apparent linear central restoring force that gives rise to the elliptical orbits. Orbits arising from a central force conserve angular momentum which means the orbit sweeps out equal areas in equal times. This is true whether it's an inverse-square law (Kepler) or a linear law (Hooke). Orbits with Hooke's law correspond to simple harmonic motion.
Obviously the motions for JWST are not small and the family of halo orbits does incude some very extreme examples. (These include highly elliptical orbits that pass close to the secondary body and are not centered very close to the Lagrange point.)
However, the JWST orbit seems relatively well-behaved. It is fairly well centered on L2 and it looks fairly elliptical. Admittedly it's not a very small orbit and it's a bit warped, but I would still guess that "equal areas in equal times" works pretty well.
Here is a plot of the triangular areas for each 1-week interval and also the velocity.
It's interesting that the velocity is highest at the inner edge of the orbit (closest to Earth). But the main feature is the twice-around variation with the velocity lowest at each y-axis extremum. The velocity variation is almost 2:1, whereas the area per unit time variation is only 4:5. So most of the velocity variation is cancelled by the radial variation - though not to the extent I was expecting.
To clarify: the first graph shows the area of the triangle $(0,0,0),r_1,r_2$ where $r_1$ is the 3D vector position at the beginning of a week and $r_2$ at the end of the week. The second graph shows the speed $|r_2-r_1|$. The vectors include x,y, and z.
[update per @BradV comment] Moving the center can get rid of the once-around component in the swept-area graph:
[update per @BradV comment #2] Moving the center to the centroid of the area enclosed by the orbit doesn't eliminate the once-around component quite as well. It doesn't take into account the velocity round the orbit (faster on the near-earth side).