Is MAX CUT approximation resistant?
If you measure approximation as a ratio between the number of constraints satisfied by your algorithm divided by the total number of constraints, then trivially all constraint satisfaction problems are unconditionally approximation-resistant.

By definition, a problem is approximation resistant if the (worst-case) approximation ratio of a random solution is (up to an additive $o(1)$ term) best possible among the (worst-case) approximation ratios achieved by all polynomial time algorithms. 

With your definition of approximation ratio, you get approximation resistance of Max Cut (and all constraint satisfaction problems)  just because you can always construct instances for which the ratio given (on average) by a random assignment is more or less (up to a $o(1)$ term) the same as the ratio given by an optimal assignment. 

For example, in the Max Cut problem, a clique on $n$ vertices is a graph that has ${n \choose 2} = n\cdot (n-1)/2$ edges, and in which an optimal cut cuts $n^2/4$ edges. This means that every algorithm, and in particular every polynomial time algorithm, has a worst-case approximation ratio (according to your definition) that is at most $1/2 + O(1/n)$ on $n$-vertex graphs. The random assignment has ratio $1/2$ on all instances, and so no algorithm can do better than the random algorithm, and the problem is approximation-resistant.