It shouldn't make a difference whether the hidden variables are generated by a computable or non-computable rule, so yes, Bell's proof should rule out non-computable local hidden variables too. Here's a simple toy model of Bell inequality violations I wrote up a while ago:
Suppose we have a machine that generates pairs of scratch lotto cards, each of which has three boxes that, when scratched, can reveal either a cherry or a lemon. We give one card to Alice and one to Bob, and each scratches only one of the three boxes. When we repeat this many times, we find that whenever they both pick the same box to scratch, they always get the same result--if Bob scratches box A and finds a cherry, and Alice scratches box A on her card, she's guaranteed to find a cherry too.
Classically, we might explain this by supposing that there is definitely either a cherry or a lemon in each box, even though we don't reveal it until we scratch it, and that the machine prints pairs of cards in such a way that the "hidden" fruit in a given box of one card always matches the hidden fruit in the same box of the other card. If we represent cherries as + and lemons as -, so that a B+ card would represent one where box B's hidden fruit is a cherry, then the classical assumption is that each card's +'s and -'s are the same as the other--if the first card was created with hidden fruits A+,B+,C-, then the other card must also have been created with the hidden fruits A+,B+,C-.
The problem is that if this were true, it would force you to the conclusion that if Alice and Bob are picking randomly which box to scratch on each trial (with a 1/3 chance of A, B, or C each time), then if they do this a large number of times, we should expect that in the subset of trials where Alice and Bob happened to pick different boxes to scratch, they should find the same fruit at least 1/3 of the time. For example, if we imagine Bob and Alice's cards each have the hidden fruits A+,B-,C+, then we can look at each possible way that Alice and Bob can randomly choose different boxes to scratch, and what the results would be:
Bob picks A, Alice picks B: opposite results (Bob gets a cherry, Alice gets a lemon)
Bob picks A, Alice picks C: same results (Bob gets a cherry, Alice gets a cherry)
Bob picks B, Alice picks A: opposite (Bob gets a lemon, Alice gets a cherry)
Bob picks B, Alice picks C: opposite results (Bob gets a lemon, Alice gets a cherry)
Bob picks C, Alice picks A: same results (Bob gets a cherry, Alice gets a cherry)
Bob picks C, Alice picks picks B: opposite results (Bob gets a cherry, Alice gets a lemon)
In this case, you can see that that if they are equally likely to pick each combination of boxes, then 2 times out of 6 when they choose different boxes, they will get the same fruit (i.e. a 1/3 chance of the same result). You'd get the same answer if you assumed any other preexisting state where there are two fruits of one type and one of the other, like A+,B+,C- or A+,B-,C-. On the other hand, if you assume a state where each card has the same fruit behind all three boxes, so either they're both getting A+,B+,C+ or they're both getting A-,B-,C-, then of course even if Alice and Bob pick different boxes to scratch they're guaranteed to get the same fruits with probability 1. So if you imagine that when multiple pairs of cards are generated by the machine, some fraction of pairs are created in inhomogoneous preexisting states like A+,B-,C- while other pairs are created in homogoneous preexisting states like A+,B+,C+, then the probability of getting the same fruits when you scratch different boxes should be somewhere between 1/3 and 1. 1/3 is the lower bound, though--even if 100% of all the pairs were created in inhomogoneous preexisting states, it wouldn't make sense for you to get the same answers in less than 1/3 of trials where you scratch different boxes, provided you assume that each card has such a preexisting state with "hidden fruits" in each box.
But now suppose Alice and Bob look at all the trials where they picked different boxes, and found that they only got the same fruits 1/4 of the time! That would be the violation of Bell's inequality, and something equivalent actually can happen when you measure the spin of entangled photons along one of three different possible axes. So in this example, it seems we can't resolve the mystery by just assuming the machine creates two cards with definite "hidden fruits" behind each box, such that the two cards always have the same fruits in a given box.
Something mathematically analogous is predicted in certain experiments where entangled photons are passed through polarizers at different angles, in which the probability that both photons will have the same result (both pass through their respective polarizers, or both are blocked) is $ cos^2 (\theta) $, where $ \theta $ is the angle between the two polarizers. Suppose the experimenters decide in advance that on each trial, they will choose randomly between one of three angles for their polarizer: 0 degrees from vertical, 60 degrees, or 120 degrees. On a given trial, if both experimenters choose the same angle, then $ \theta = 0 $ so they are guaranteed to get the same result with probability 1 (both photons pass or both are blocked), but if they choose different settings the probability of getting the same result would be $ cos^2 (\pm 60) $ or $ cos^2 (\pm 120) $, which in both cases gives a probability of 1/4. But the Bell inequality whose derivation I sketched says that if you want to explain the perfect match when both experimenters make the same choice using local hidden variables, the probability of a match when they make different choices should be no lower than 1/3.