Can we solve the Hidden Number Problem in $GF(2^n)$?
Yes, it appears that it can be solved in practical time in $GF(2^n)$, if the attacker gets $n+\epsilon$ random $a_i$ values, even if he gets a single bit of the $a_i \times k$ values.

The chief observation is that the mapping from $a_i$ to bit $j$ of $a_i \times k$ (which I'll refer to as $bit_j(a_i \times k)$) is bitwise linear (for constant $j$, $k$).

So, assuming that we have $a_i$ and the corresponding $bit_m(a_i \times k)$ values (for some known $m$), our first step is to find $n$ linear independent $a_i$ values (which, given $n+\epsilon$ random $a_i$ values, we'll have with good probability); we view this as a set of $n$ linear equations, and solve them in $O(n^3)$ time, giving us the mapping between any value $a$ and $bit_m(a \times k)$

Now, for polynomial representations of $GF(2^n)$, for every $i, j$ pair, there is a constant $c_{i,j}$ where $bit_i(a) = bit_j(a \times c_{i,j})$ for all $a$.

This observation allows us to find the linear mapping of $bit_n(a \times k)$ for any $n$, as $bit_n(a \times k) = bit_m(a \times k \times c_{n,m}) = bit_m( (a \times c_{n,m} ) \times k )$; in other words, to perform this mapping, we multiply the vector by the constant $c_{n,m}$, and then apply the mapping we already know (both of which are linear operations).

The result of this are $n$ linear maps giving us all $n$ bits of the product.  Then, we can solve these $n$ linear maps for the input that gives a 1 bit for the lsbit map, and 0 for the rest; the solution for this is $k^{-1}$, which immediately gives us the value for $k$.

I suspect that there are signficantly more efficient ways to solve this; however this algorithm should be enough to answer the question.