When studying such problem, cryptographers assimilate a cryptographic hash $H$ to a random function or random oracle (or random mapping though that's a bit dated) that's iterated. Probabilities are computed over the set of all possible hashes. That's an approximation, but if actual results where distinguishably different, that would be a way to distinguish the hash from a random function, thus a break of the hash by modern defintions. Therefore, said approximation is satisfactory, and the better there can be for an unbroken cryptographic hash.
Will there be short cycles?
Likely, and the more so as we loosen the definition of short. But (except for very small hashes) it's unlikely that a short cycle would be reached from a random starting point $A$; and (for standard cryptographic hashes with large enough output that they are collision-resistant) it's unlikely we could exhibit any cycle, regardless of it's size.
Are there values of $A$ such that $H(A)=A$ (a cycle of 1)
If the destination set of the hash has size $n$ (where $n=2^b$ for a $b$-bit hash, e.g. $n=2^{256}$ for SHA-256), then the probability that there is a cycle of size 1 is easily computed as the complement of the probability that none of the $n$ points hashes to itself, that is $p_1(n)=1-(1-1/n)^n$. This starts from $p_1(1)=1$, $p_1(2)=3/4=0.75$, $p_1(3)=19/27\approx0.7037$, and $p_1$ quickly converges to $1-1/e\approx0.6321$.
Thus there's >63.2% probability there exists a cycle of size 1 in a given cryptographic hash like SHA-256, SHA-384, or SHA-512. And we can't tell better for any of these hashes.
I don't know how to properly do the same for cycles of size 2, but do not doubt it is feasible, and the probability is sizable for $n\ge2$.
It's possible to compute the expected value of many characteristics of cycles for an iterated random function/hash. In particular, for large $n$, the expected number of steps before hitting a previous value starting from a random point is $\approx\sqrt{\pi\,n/2}$, and the expected length of the cycle reached is half that. A classical reference is Philippe Flajolet and Andrew M. Odlyzko, Random mapping statistics, in proceedings of Eurocrypt 1989, and Research Report RR-1114, INRIA, 1989.
Is there a way to prove that either no such cycles exist for a given hash, or to put a lower bound on the shortest cycle.
No, for an unbroken cryptographic hash of large enough output size that it's collision resistant (that is, $\sqrt n$ is so large that this number of hashes can't be computed); say, any unbroken hash 256-bit or wider. For the part of the question asking is it's possible to prove that no such cycle exist, we'd even need to compute $n$ hashes (until the last we can't be sure there's not a cycle of size 1), thus any unbroken hash 128-bit or wider will do.
So how does (Douglas Hofstadter's iterated construction) connect to cryptography?
A quite similar technique is used in attacking some cryptosystems. We construct a function, iterate it until finding a cycle (which thus is short, else we could not find it), and the entry point in the cycle gives a collision, and the collision solves the problem because the function was constructed with that explicit purpose. That's the heart of Pollard's rho method to solve the Discrete Logarithm Problem, which is the best theoretical attack we have for this problem in some groups used in cryptography.
Notice that neither the function in the above, nor the function constructed by Douglas Hofstadter, constitutes a secure cryptographic hash. It's because they are not that we can find a cycle, and solve the problem at hand.