Does there exist a semi-prime integer $n$ (e.g., an RSA modulus) such that the order of every element in the multiplicative group modulo $n$ is equal to the order of the full group modulo $n$?
No, but we can come close; if we gave the product of two distinct safeprimes $n = (2r+1)(2s+1)$ (where $r, s$ are prime), then almost all the elements in $\mathbb{Z}_n^*$ are either of order $rs$ or of order $2rs$. The exceptions are the values that happen to be 1, -1 modulo $2r+1$ or $2s+1$ - if $r, s$ are large, such values are rare (and if you happen to stumble on one of those values (other than 1, n-1), you can factor anyways).
On the other hand:
What I am thinking here is that if the number of qubits required to factor an RSA modulus is directly proportional to the order of a random element selected from the group
This wouldn't help that much; if we consider the order of a random element of $Z_n^*$, the order will be $n/d$ for a relatively small $d$; exceptions will be comparatively rare (and someone who doesn't know the factorization won't be able to produce them without guessing).
Here's how it works: the order of a value $g$ modulo $n$ is the least common multiple of the order of $g$ modulo $p$ and the order of $g$ modulo $q$ - hence, we need to look at the behavior modulo those primes to understand how the order modulo $n$ works. Now, if $r$ is a divisor of $p-1$, then the probability that a random $g$ has an order which is a divisor of $(p-1)/r$ is $1/r$ - hence if $r$ is large, this happens only with small probability (and if $r$ is small, this doesn't reduce the order much).
Now, one case where the order might be smaller if the value of $\gcd(p-1, q-1)$ is large (because that is effectively a division of the least common multiple). However, with random $p, q$, this is usually small; it's not clear whether going to extra effort to make sure that it is minimal (2, which it is in the clear where $p, q$ are safeprimes) is worth the effort.
