OK, assuming a setup whereby even if the location of the $k$ bits are unknown the guess or is told by an honest party whether his guess is correct, one can say the following.
Define a collision resistant hash with $n$ bit output as one which requires $O(2^{n/2})$ guesses before finding a collision.
If there was a way of finding a collision in an arbitrary $k$ bit subset in time strictly less than $O(2^{k/2})$ the the overall time would be reduced by attacking, say the $k$ bits and the remaining $n-k$ bits separately (in time strictly less than $O(2^{(n-k)/2})$) getting an overall attack time strictly less than $O(2^{n/2}).$
If $k$ needed to be bounded by some function of $n$ you could get the same result by partitioning into $\lceil n/k\rceil$ subblocks of length $k$ each.
If the multiplicative speedup factor for a k bit block was $2^{-\theta k}$ with $\theta \in (0,1),$ your overall speedup would be
$$
2^{-\lfloor n/k \rfloor \theta k}
$$
approaching
$$
2^{- \theta n}.
$$
If the weakness was local for a specific $k$ bits, the saving would only be
$$2^{- \theta k}$$
but still contradict the definition of collision resistant $n$ bit hash function.
