Let me try to work this out, along the lines of a <A HREF="http://mathoverflow.net/questions/194840/some-calculus-in-the-orthogonal-group-on/194900#194900">similar calculation</A> 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 <A HREF="http://statweb.stanford.edu/~cgates/PERSI/papers/functionals.pdf">Diaconis and Evans</A>:
$$n^2 a_m(n)+nb_m(n)+nc_m(n)=\int_{{\rm U}(n)}\,({\rm tr}\,U^m)({\rm tr}\,\bar{U}^m)\,dU={\rm min}\,(n,m)$$
*[note: in a related <A HREF="http://mathoverflow.net/questions/195186/expectation-of-trace-of-nth-power-of-unitary-matrices">MO posting</A> 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
$$na_m(n)+nb_m(n)+n^2 c_m(n)=\int_{{\rm U}(n)}\,{\rm tr}\,(U^m\bar{U}^{m})\,dU$$

To evaluate this integral I write $U^m=V\Lambda^m V^\dagger$, with $\Lambda$ the diagonal matrix of eigenvalues of $U$ and $V\in{\rm U}(n)$ independent of $\Lambda$. I then first average over the $V$ matrices, which is easy because there are just four of them, and then the remaining average over $\Lambda$ can be evaluated using Diaconis and Evans:

$$\int_{{\rm U}(n)}\,{\rm tr}\,(U^m\bar{U}^{m})\,dU =\frac{1}{n+1}+\frac{1}{n(n+1)}\int_{{\rm U}(n)}\,({\rm tr}\,U^m)({\rm tr}\,\bar{U}^{m})\,dU=\frac{n+{\rm min}\,(n,m)}{n(n+1)}$$

so now we have three equations with three unknowns and we're done:

$$a_m(n)= \frac{\min(n,m)  \left(n (n+1)^2-1\right)-n \left(n^2+n+1\right)}{(n-1) n^2 (n+1) (n+2)}$$
$$b_m(n)= \frac{n \left(n (n+1)^2-1\right)-\min(n,m)  \left(n^2+n+1\right)}{(n-1) n^2 (n+1) (n+2)}$$
$$c_m(n)= -\frac{\min(n,m) +n}{n^2 (n+2)}$$