This is by no means a comprehensive answer on this subject, but perhaps it's a good start.
Shor's algorithm for (specific) ECC
This paper by Proos and Zalka compares implementations of Shor's algorithm for integer factorization and discrete logarithms for some elliptic curve groups (notably, ONLY those over prime finite fields).
If $n$ is the bit length of the group (from section 6.3):
RSA | ECC
n | qubits | time | n | qubits | time
1024 | 2048 | 4.3*10^9 | 163 | 1000 | 1.6*10^9
2048 | 4098 | 34*10^9 | 224 | 1300 | 4*10^9
3072 | 6144 | 120*10^9 | 256 | 1500 | 6*10^9
RSA needs about $2n$ qubits, while ECC needs approximately $6n$ (for smaller $n$). Similarly, RSA needs ~$4n^3$ gates, while ECC only requires ~$360n^3$.
ECC is classically more difficult than RSA, allowing for smaller key sizes and more efficient computation. Quantumly, ECC is still more difficult than RSA, but less so, leading to what the authors call a "quantum advantage" for elliptic curve discrete logarithms over integer factorization.
Comparing RSA-3072 to ECC-256, they found that a quantum computer needs to be approximately $4\times$ as big and takes approximately $20\times$ as much time to crack RSA as compared to ECC.
Keep in mind that this work is from 2008 and is only relevant for curves over prime fields.
Obviously a factor of 4 in the number of qubits is not "much more quantum-resistant" as per the link given by fgrieu. It's very possible that ECC over GF($2^m$) is much more efficiently broken by quantum computing.
Due to the similarities of the two problems, I can't imagine that qubit configuration would have much of an impact in one direction or the other.
However, this is all conjecture on my part.
It is not likely that the gap between quantum
cryptanalysis of a 384-bit key and a 3072-bit key will be great enough to
serve as a basis for a cryptographic strategy.
In any case, I interpret the above quote from as there being no practical difference between the two. Both should be considered utterly broken at that time.