A TSP to HamCycle Reduction
By direct reduction, do you then mean a reduction that does not leave the graph-world and go back again? (that might be a very vague formulation, sorry). I do not think such a reduction exists. Of cause proving that it does not exist is not a simple matter, but at least I have not been able to find or think of one myself. But here's a couple of thoughts about why such a direct reduction is unlikely:

TSP is run on a fully connected graph with $n$ nodes and $n^2$ edges while HamCycle is run on a graph with $n$ nodes and $m &lt; n^2$ edges (otherwise it would be trivial). 
A solution to TSP must pass all nodes and contain no cycles and therefore must contain exactly $n$ edges. We could run HamCycle each on all possible paths of length $n$, starting from the one with the lowest edge-sum and continuing until we reach one that contains a Hamiltonian Cycle, but as there are ${{n^2}\choose{n}} = O(n^{2n})$ such possible paths, this would mean running HamCycle an exponential number of times. I can not see how you can avoid this.

You have most likely thought about this already, so this is mostly to say that you are not the only one who are not able to come up with such a reduction and I suspect that this may be because a reduction either has to be quite complicated or has to leave the graph-world somehow and go back again. Of cause there might exists a reduction that does leave the graph-world but is still simpler than going all the way through SAT, but I do not know of any such reduction. 

I'm quite interested to see if anyone has a better answer to this.