For the multiple pattern case, it seems that simply scanning for each of the might be the best possible solution, at least unless the strong exponential-time hypothesis fails.
Recall that given sets $S_1, S_2, \dotsc, S_n$ and $T_1, T_2, \dotsc, T_n$ over universe $[m]$, if we could decide if there are $S_i$ and $T_j$ such that $S_i \cup T_j = [m]$ in time $O(n^{2-\varepsilon}\operatorname{poly}(m))$, then SETH fails, i.e. we have a CNF-SAT algorithm with running time $O^*\bigl(2^{(1-\varepsilon/2)n}\bigr)$.
Given sets $S_1, S_2, \dotsc, S_n$ and $T_1, T_2, \dotsc, T_n$, we encode the above problem as multi-pattern matching with don't cares over binary alphabet as follows:
- The text is $$1[T_1]10^{m+2}1[T_2]10^{m+2}\dots0^{m+2}1[T_n]1\,,$$ where $[T_i]$ is the natural encoding of $T_i$ as a binary string.
- We have $n$ patterns of form $1\langle S_i \rangle 1$, where $\langle S_i \rangle$ is a string $y = y_1y_2\dots y_m$ such that $y_j = 1$ if $j \notin S_i$ and $y_j = * $ if $j \in S_i$ (here $*$ is the don't care symbol).
Now it's clear that a pattern $1\langle S_i \rangle 1$ can match the text at an occurrence of $1[T_j]1$, and only when $S_i \cup T_j = [m]$. The total length of patterns and the length of the text are both $O(nm)$, for instance so a near-linear single-pass algorithm for multiple patterns would give substantial improvements over best known CNF-SAT algorithms...
(Note that this does not say anything about algorithms that use lots of time preprocessing the patterns, say, quadratic in the total length of the patterns.)