I am not exactly sure if I understand your question correctly. I will assume that by thermal superconductivity you mean perfect conduction of heat.

This is indeed an interesting question: Are there systems that are analogous to superfluids and superconductors, perfect conductors of particle number and charge, that are perfect conductors of heat?

I think the answer is no. In superfluids and superconductors there is a spontaneoulsy broken symmetry, and an associated Goldstone boson $\varphi$. Gradients of $\varphi$ defines a superfluid velocity that enters in the hydrodynamic description and describes transport without dissipations. In a superfluid, $v_s=\nabla\varphi/m$ is the superfluid velocity. The equation of motion $\partial_tv_s =-\nabla\mu+\ldots$ shows that an arbitrarily small gradient of the chemical potential will drive a flow. Gradients of $v_s$ do not contribute to the stress tensor, so there is no dissipation. In a superconductor we have the London current $j_s=e^2n/(mc)(\nabla\varphi-A)$, and the equation of motion is $\partial_t(\nabla\varphi)=-\nabla V+\ldots$. The London current does not contribute to Ohmic heating.

So can we find a Goldstone boson that contributes to the energy current, and has an equation of motion of the form $\partial_t(\nabla\varphi)=-\nabla T+\ldots$? I see several difficulties with this: The symmetry would have to be time reparametrization invariance (or Lorentz boosts in the relativistic case). Also, $\nabla T$ does not appear in the effective Lagrangian (which is a $T=0$ object), it would have to arise from integrating out thermal fluctuations (but that will usually lead to dissipative terms, because of fluctuation-dissipation relations). This is not a proof, of course, but at least it shows that it is not obvious how to obtain perfect conductors of heat.

There are systems that are extremely good conductors of heat:

1) In a typical material small gradients of T do not lead to convection, the heat current is diffusive and proportional to $-\kappa\nabla T$. This is how we measure $\kappa$. In superfluids (like liquid He) small gradients of T drive a new type of convection, involving a flow of the normal component balanced by a backflow of the superfluid. The energy current is not proportional to $-\kappa\nabla T$, so the thermal conductivity appears to be very large. Helium does, however, have a finite thermal conductivity which can be measured by attenuation of first and second sound.

2) There are systems in which the mean free path (of electrons or phonons or other carriers) is larger than the system size. The usual example is nanowires or carbon nanotubes. Transport is ballistic, not diffusive, and a naive measurement of thermal conductivity will give a very large $\kappa$. There is a large amount of literature on this, look for the Landauer formula or the Landauer-Buttiker formalism.