What will be formed when concentrated hydrochloric reacts with copper oxide?

If, by “copper oxide”, you mean CuO, then:

2CuO(a) + 2HCl(a) = 2CuCl(a) + H2O + 0.5O2(g) [T > 35C]

Change in Free Energy: ΔG(35C) = -1.9kJ (negative, so the reaction runs)

Change in Enthalpy: ΔH(35C) = +132.1kJ (positive, so the reaction is endothermic)

However, if, by “copper oxide”, you mean Cu2O, then:

Cu2O + 2HCl(a) = 2CuCl(a) + H2O

Change in Free Energy: ΔG(35C) = -23.4kJ (negative, so the reaction runs)

Change in Enthalpy: ΔH(35C) = +16.6kJ (positive, so the reaction is endothermic)

BUT if T > 195C, then

Change in Free Energy: ΔG(195C) = -44.0kJ (negative, so the reaction runs)

Change in Enthalpy: ΔH(195C) = -0.012kJ (negative, so the reaction is exothermic)

And the reaction becomes more exothermic as the temperature is increased beyond 195C.

Copper chloride and water are formed.
The equations (for copper(I) and copper(II) oxides) are:
Cu₂O + 2HCl → 2CuCl + H₂O.
CuO + 2HCl → CuCl₂ + H₂O.