What is the thermal state of a reaction at equilibrium?
You're right, as far as I can see. If we look at a particular transformation $A \leftrightharpoons B$ there is a forward reaction and a backwards reaction. They are associated with enthalpy change of different signs and the same magnitude. At equilibrium the rates are the same, so the forwards and backwards reactions release and absorb the same amount of heat and it appears as if nothing is happening.

There is no separate "reaction at equilibrium" with its own enthalpy change. It's not really clear to me what "thermal state of a reaction" means, as systems have states and reactions are pathways between systems.