You mean an oscillation around some equilibrium orientation due to a torque, that is not dampened by some friction or radiation? That would (almost) always be harmonic for sufficiently small amplitudes by approximation.
One exception would be particular places where the Taylor series of torque as a function of the angle has no linear term ~ $\alpha$, like $t(\alpha) = k\cdot\alpha^3$. That would not yield a harmonic motion, and I don't know a general answer for that case - one would need to study non-linear ordinary differential equations.
Anyway, for larger amplitudes of $\alpha$, it would depend on how exactly the torque increases with the angle. As soon as the torque back to equilibrium deviates from a linear=proportional increase with the angle, the resulting motion cannot be harmonic (sine-shaped) any more:
Harmonic motion like $\alpha(t) = \alpha_0\cdot\sin(t)$ means that the second derivative of $\alpha$ is sinusoidal, too, but with a negative sign, just derive it two times! But this angular acceleration $\ddot{\alpha}$ is always proportional to the torque, analogously to a ~ F with linear motion according to Newton. Thus, the torque needs to increase proportionally to the negative angle (since both follow the sine in time) - or else there is no perfect harmonic motion in the first place.
But even in a homogeneous/uniform field, the torque increases with the sine of the angle, so not linearly for large angles, and it is only approximately linear for small angles where $sin(\alpha) \approx \alpha$. Think of two opposite point charges at the end of a stick. The torque only counts the tangential components of the otherwise constant forces on the charges. It is analogous to the gravity pendulum that also displays true harmonic motion for sufficiently small angles only. The dipole behaves exactly like that pendulum up to 90 degrees.