Derangement formula for repeated permutation

Yes, there is a formula for counting these generalized derangements. It is due to Even and Gillis and it is in terms of products of Laguerre polynomials. See e.g. this paper by Gessel for a derivation from rook theory (p.4). Let $$l_n(x) = \sum_{k=0}^n (-1)^k { n \choose k}^2 k! x^{n-k},$$ and define $\Phi$ to be the linear function on polynomials mapping $x^n$ to $n!$. It's shown that $$\Phi\left( \prod_{i=1}^r l_{n_i}(x)\right)$$ is the number of permutations of $n_1 + \cdots + n_r$ colored elements, with $n_i$ of the $i$-th color, so that no element is mapped to an element of the same color.

Here all elements are labeled, so the elements in a given color class are distinguishable, but if you don't want that just divide by $\prod_{i=1}^r n_i!$ to account for the permutations of each color class. And if you want a more compact formula, note that $\Phi(p(x)) = \int_0^\infty e^{-x} p(x)\,dx$.

Edit: Here are some more details.

First, an example of using this formula. The first few polynomials $l_n(x)$ are

\begin{align*} l_0(x) &= 1\\ l_1(x) &= x - 1\\ l_2(x) &= x^2 - 4x + 2\\ l_3(x) &= x^3 - 9x^2 + 18x - 6. \end{align*}

Now let's find the number of permutations of $1234$ where $1,2$ are colored red and $3,4$ are colored blue and no element can map to another with the same color. Since there are $2$ of each color, we get $$l_2(x)l_2(x) = (x^2 - 4x + 2)(x^2 -4x + 2) = x^4 - 8x^3 + 20x^2 - 16x + 4.$$

Then we just apply $\Phi$, which means replacing each variable $x^k$ with $k!$. We get

\begin{align*} \Phi(l_2(x)l_2(x)) &= 1 \cdot 4! - 8 \cdot 3! + 20 \cdot 2! - 16 \cdot 1! + 4 \cdot 0!\\ &= 4. \end{align*}

This corresponds to the $4$ permutations $3412, 4312, 3421, 4321$ with no element of $\{1,2\}$ mapping to an element of $\{3,4\}$ or vice versa.

If you want the number of derangements of the multiset $1122$ where the $1$s and $2$s are not distinguishable, just divide take this answer and divide by $2! \cdot 2!$ to get $1$, corresponding to the single word $2211$.

As for the proof - I will not give it entirely, but I will give the main ingredients. (Note: We use $[n]$ to mean the set $\{1,2, \ldots, n\}$ and $[m,n]$ to mean $\{m, m+1, \ldots, n-1, n\}$.)

1. There is a well-known formula from rook theory, proven using inclusion-exclusion. If $B$ is a "board", a subset of the $n \times n$ grid $[n] \times [n]$, then let $r_k$ be ways of placing $k$ elements on the board $B$ with no two in the same row or column (i.e., the number of ways of placing $k$ rooks from chess that cannot attack each other.) Then $$\sum_{k} (-1)^k r_k (n-k)!$$ is the number of permutations $\sigma \in S_n$ with no $\sigma(i) = j$ for $(i,j) \in B$; that is, no $1$s on the set $B$ when you write the adjacency matrix. You can write this as $\Phi(p_B(x))$ where $$p_B(x) = \sum_{k=0}^n (-1)^k r_k x^{n-k}$$ is the "rook polynomial" for $B$. (Note this is a variant of the usual definition of rook polynomial.)

2. If $$B_1 \subseteq [n_1] \times [n_1], B_2 \subseteq [n_2]\times[n_2],$$ let $B_1 \oplus B_2$ be board in $[n_1 + n_2] \times [n_1 + n_2]$ given by the disjoint union of $B_1$ with the translate of $B_2$ to the upper-right square $[n_1+1, n_1 + n_2] \times [n_1+1, n_1 + n_2]$. Then $p_{B_1}(x) p_{B_2}(x) = p_{B_1 \oplus B_2}(x)$. Inductively we get $p_{B_1}(x) \cdots p_{B_k}(x) = p_{B_1 \oplus \cdots \oplus B_k}(x)$, the rook polynomial for the board given by the block-diagonal board $B_1 \oplus \cdots \oplus B_k$.

3. Show that if $B$ is the whole board $[n] \times [n]$ then $p_B(x) = l_n(x)$ given above.

4. Note that if we have boards $B_i = [n_i] \times [n_i]$ for some $n_i$, permutations of $[n_1 + \cdots + n_k]$ avoiding the block-diagonal $B_1 \oplus B_2 \oplus \cdots \oplus B_k$ are exactly the generalized derangements: no $i \in B_l$ can map to $j \in B_l$ for any $l$. Then we count these by applying 1, 2, 3 above.