Pretty much the title. I've heard it said that no (see, e.g here, answer by Lukas Schaedler). To quote his argumentation:
No. Exergonic reactions may be both exothermic or endothermic. Endergonic reactions are endothermic only.
Endergonic and exergonic relate to changes in free energy (delta G), while endothermic and exothermic are related to changes in enthalpy (delta H).
Gibb’s equation written in form of free energy (G), enthalpy (H) and entropy (S) changes:
delta G = delta H - T . delta S
Therefore, it’s possible for an exergonic reaction (delta G < 0; decrease in free energy) to be both exothermic (delta H < 0; heat is released) or endothermic (delta H > 0; heat is consumed). The second scenario may happen if :
T . delta S > delta H
On the other hand, endergonic reactions are endothermic only; delta H needs to be positive and greater than T . delta S in order to delta G > 0
But it's not really clear to me why. In particular, in the example given, why is $\Delta S \ge 0$ (which in my understanding would be necessary to force $\Delta G = \Delta H - T\Delta S$ to be negative)?