Let me try to work this out, along the lines of a similar calculation in the orthogonal (rather than unitary) group.
We need the fourth-order tensor $$\int_{{\rm U}(n)}(U^m)_{ij}(\bar{U}^m)_{kl}\,dU=a_{m}(n)\delta_{ij}\delta_{kl}+b_{m}(n)\delta_{ik}\delta_{jl}+c_{m}(n)\delta_{il}\delta_{jk}$$ so that the required integral takes the form $$\int_{{\rm U}(n)}U^m X(U^\dagger)^m\,dU=a_{m}(n)X+b_{m}(n)\mathbb{1}\,{\rm tr}\,X+c_{m}(n)X^{\rm t} $$ [note: The OP does not have the transpose $X^{\rm t}$, but I don't see a priori why this term will not appear.]
Substitution $X=\mathbb{1}$ gives a first relation $$a_m(n)+nb_m(n)+c_m(n)=1$$ one more relation follows from application of theorem 2.1.b of Diaconis and Evans: $$n^2 a_m(n)+nb_m(n)+nc_m(n)=\int_{{\rm U}(n)}\,({\rm tr}\,U^m)({\rm tr}\,\bar{U}^m)\,dU={\rm min}\,(n,m)$$ [note: in a related MO posting I had $m$ instead of ${\rm min}\,(n,m)$, I have now corrected this oversight and apologize for the confusion it may have caused]
I need a third relation $$na_m(n)+nb_m(n)+n^2 c_m(n)=\int_{{\rm U}(n)}\,{\rm tr}\,(U^m\bar{U}^{m})\,dU$$ This third integral remains to be evaluated, I will do that later today.
as an intermezzo, suppose as in the OP that there is no term $X^{\rm t}$, so $c_{m}=0$. Then $nb_m=1-a_m$ and $a_m= [{\rm min}\,(n,m)-1](n^2-1)^{-1}\equiv p\in[0,1]$ so the inconsistency noted by the OP does not appear; still, I'm not sure why $c_m$ would vanish.