Are all central forces conservative?

Depends on what you mean by 'central force'.

If your central force is of the form ${\vec F} = f(r){\hat r}$ (the force points radially inward/outward and its magnitude depends only on the distance from the center), then it is easy to show that $\phi = - \int dr f(r)$ is a potential field for the force and generates the force. This is usually what I see people mean when they say "central force."

If, however, you just mean that the force points radially inward/outward, but can depend on the other coordinates, then you have ${\vec F} = f(r,\theta,\phi){\hat r}$, and you're going to run into problems finding the potential, because you need $f = - \frac{\partial V}{\partial r}$, but you will also need to have $\frac{\partial V}{\partial \theta} = \frac{\partial V}{\partial \phi} = 0$ to kill the non-radial components, and this will lead to contradictions.

It's logical that a field of this form is gong to be nonconservative, because if the force is greater at $\theta = 0$ than it is at $\theta = \pi/2$, then you can do net work around a closed curve by moving outward from $r_{1}$ to $r_{2}$ at $\theta = 0$ (positive work), then staying at $r_{2}$ constant, going from $\theta =0 $ to $\theta = \pi/2$ (zero work--radial force), going back to $r_{1}$ (less work than the first step), and returning to $\theta = 0$ (zero work).

Take curl in spherical polar coordinates of a central force ,you will see that since there is no component of the force in the $\theta$ and $\phi$ direction and $f(r)$ doesn't depend on $\theta$ and $\phi$ ,so curl of central force is zero. Hence central forces can be represented as gradient of some scalar,i.e. central forces are conservative.