Could min cut be easier than network flow? Thanks to the max-flow min-cut theorem, we know that we can use any algorithm to compute a maximum flow in a network graph to compute a $(s,t)$-min-cut.  Therefore, the complexity of computing a minimum $(s,t)$-cut is no more than the complexity of computing a maximum $(s,t)$-flow.

Could it be less?  Could there be an algorithm for computing a minimum $(s,t)$-cut that is faster than any max-flow algorithm?

I tried finding a reduction to reduce the $(s,t$)-max-flow problem to the $(s,t)$-min-cut problem, but I wasn't able to find one.  My first thought was to use a divide-and-conquer algorithm: first find a min-cut, which separates the graph into two parts; now recursively find a max-flow for the left part and a max-flow for the right part, and combine them together with all of the edges crossing the cut.  This would indeed work to produce a maximum flow, but its worst-case running time might be as much as $O(|V|)$ times as large as the running time of the min-cut algorithm.  Is there a better reduction?

I realize the max-flow min-cut theorem shows that the complexity of computing the value of a max-flow is the same as the complexity of computing the capacity of a min-cut, but that's not what I'm asking about.  I'm asking about the problem of finding a max-flow and finding a min-cut (explicitly).

This is very closely related to Compute a max-flow from a min-cut, except: (1) I'm willing to allow Cook reductions (Turing reductions), not just Karp reductions (many-one reductions), and (2) perhaps given $G$ we can find some graph $G'$ such that the min-cut of $G'$ makes it easy to compute the max-flow of $G$, which is something that's out of scope for that other question.
Here is a possible approach:

Suppose you know the cut S, then finding the flow from $S$ to $t$ is a min cost network flow problem with zero cost, since you know exactly the outflow at each vertex in $V\setminus S$ and in the inflow at $t$. Suppose $f$ denotes an $S-t$ flow and $A$ the node-arc matrix (i.e. row $i$, col $j$ has 1 if $i$ is the tail of $j$, -1 if its the head, zero otherwise), and let $b$ be such that $Af = b$ if $f$ satistfies the supply/demand, and the flow conservation. Then with gaussian elimination we can find a feasible solution in $|V|^3$ operations. 

To find a cut from a flow we need to construct the residual graph which takes at most $|E|$ time, and then potentially traverse $|V|$ vertices.  

So for complete graphs and the minimum cut being just the source or just the sink, the reduction takes equal time in both directions worst case. However, I would think that finding a feasible solution to $Af =b$ can be done faster than $|V|^3$ given the special structure. I am not sure how to prove that though.