Using your posted A matrix I find that
For numerical purposes, this is a singular matrix and shouldn't be treated as positive definite (but rather positive semi-definite). The test you did with chol() isn't reproducible,
>> [~,p] = chol(H); p
but even if p had been 0, it would be irrelevant. chol's judgement on whether a matrix is sufficiently positive definite to compute a Cholesky decomposition does not mean it is sufficiently positive definite for everything.
For quadprog, the only thing that matters here is that, due to the ill-conditioning of H, the equations H*x+f=0 will have a non-unique set of solutions that all approximately minimize the quadprog objective. You simply cannot be sure which of these solutions different algorithms will produce, with such an ill-conditioned H.
As for why the residual is so much worse for quadprog, that is likely because quadprog is an iterative solver and cannot use the residual to decide when to stop iterating (because it doesn't have access to A and B). In fact, it almost certainly uses something like a stopping tolerance on the gradient norm,
gradnorm = norm(H*x+f)^2 < tolerance
Since H*x+f=2*A.'*(A*x-B), we can see that the gradient norm can be significantly smaller than the residual vector r= A*x-B depending on the values of A.'