This is an extended version of my comment of Weingarten functions, a brute force computation that perhaps not very insightful.

I'll call $m+ k = n$ in the following and use $\dagger$ to denote Hermitian conjugate.

Let $\{e_i\}$ to be a basis in $\mathcal{H}$ then the basis in $\mathcal{H}^{\otimes m} \times \mathcal{H}^{\otimes n}$(where $E$ lives in) is
\begin{equation}
e_{i_1} \otimes e_{i_2} \otimes \cdots \otimes e_{i_m} \otimes e^{\dagger}_{j_1} \otimes e^{\dagger}_{j_2} \otimes \cdots \otimes e^{\dagger}_{j_n}
\end{equation}
The matrix $E$ can be decomposed as follows where repeated indices are assumed to be summed over
\begin{equation}
E = E_{i_1 i_2 \cdots i_m , j_1 j_2 \cdots j_n} e_{i_1} \otimes e_{i_2} \otimes \cdots \otimes e_{i_m} \otimes e^{\dagger}_{j_1} \otimes e^{\dagger}_{j_2} \otimes \cdots \otimes e^{\dagger}_{j_n}
\end{equation}
Similarly $U_{ij}$ are the components of $U$ under $e_i \otimes e_j^\dagger$ basis, then the integral becomes
\begin{equation}
\begin{aligned}
\int_U U^{\dagger}_{\bar{i}_1 i_1} U^{\dagger}_{\bar{i}_2 i_2} \cdots U^{\dagger}_{\bar{i}_{n+m} i_{n+m} }U_{ j_1 \bar{j}_1 } U_{ j_2 \bar{j}_2 } \cdots U_{j_{n+m}\bar{j}_{n+m} } \\
E_{i_1 i_2 \cdots i_m,j_1 j_2 \cdots j_n} X_{\bar{j}_1 \bar{j}_2 \cdots \bar{j}_n \bar{i}_{m+1} \cdots \bar{i}_{m+n}} E^{\dagger}_{i_{m+1} i_{m+2} \cdots i_{m+n} ,j_{n+1} j_{n+2} \cdots j_{n+m}} dU
\end{aligned}
\end{equation}
the free indices are $\bar{i}_1, \bar{i}_2, \cdots, \bar{i}_m$ and $\bar{j}_{n+1}, \bar{j}_{n+2}, \cdots \bar{j}_{n+m}$.

The integral of the component of unitary matrices can be represented as a series over all possible permutations $\tau, \sigma$
\begin{equation}
\int_{\mathcal{U}(N)} \prod_{k=1}^{n} U^{\dagger}_{\bar{i_k} i_k} U_{j_k \bar{j}_k} dU=\sum_{\tau,\sigma\in S_n} {\rm Wg}^U(\tau^{-1}\sigma)\prod_{k=1}^n \delta_{i_k,\tau(j_k)}\delta_{\bar{i}_k,\sigma(\bar{j} _k)}
\end{equation}
where $\rm Wg$ is the Weingarten function.

In this summation, the indices with and without bar are contracted separated, so $E$ will contract with $E^{\dagger}$ for different permutations and $X$ will contract with itself.

To further simplify this integral, you will have to explore the properties of the Weingarten function. For example when $\sigma = \tau$, the function is $1$ and all the others will be of order $\mathcal{O}(\frac{1}{d})$. If $d$ is a large number this will give you the leading order contribution.

Even the $m = 1$ case is hard. When $n > 1$, the contractions can't be expressed as trace or matrix product. There are in general $(n+m)!$ terms and I don't know how to obtain a closed form expression.

Let me demonstrate how to compute $n = 1$(which may not be trivial to OP). In that case we have $E_{i_1 j_1} X_{\bar{j}_1 \bar{i}_2 } E^{\dagger}_{i_2 j_2}$ contracted those deltas. $n+m = 2$, so there are only two possible permutations for $\sigma $ and $\tau$. Check out the Wikipedia table, when they are the same ${\rm Wg} = \frac{1}{d}$, when they are different ${\rm Wg} = \frac{-1}{d(d^2 -1)}$. Therefore the integral is
\begin{equation}
\begin{aligned}
\frac{1}{d} \left[ E_{i_1 i_1} E^{\dagger}_{i_2 i_2 } X_{\bar{i}_1 \bar{j}_2 }
+ E_{i_1 i_2 }E^{\dagger}_{i_2 i_1} X_{\bar{i}_2 \bar{i}_2 } \delta_{\bar{i}_1 \bar{j}_2 } \right]\\
- \frac{1}{d(d^2 - 1 ) } \left[ E_{i_1 i_1} E^{\dagger}_{i_2 i_2 } X_{\bar{i}_2 \bar{i}_2 } \delta_{\bar{i}_1 \bar{j}_2 } + E_{i_1 i_2 }E^{\dagger}_{i_2 i_1} X_{\bar{i}_1 \bar{j}_2 } \right]
\end{aligned}
\end{equation}

In the matrix notation it is
\begin{equation}
\begin{aligned}
\frac{1}{d} \left[ {\rm tr}(E) {\rm tr}(E^{\dagger}) X
+ {\rm tr}(E E^{\dagger}) {\rm tr}( X) I \right]
- \frac{1}{d(d^2 - 1 ) } \left[ {\rm tr}(E) {\rm tr}(E^{\dagger}) {\rm tr}( X) I + {\rm tr}(E E^{\dagger}) X \right]
\end{aligned}
\end{equation}