If you take the diagram of the Reidemeister 3 move and "shortcircuit" two ends, you get (click http://imgur.com/kRvZa if Imgur hotlink doesn't work):
I have circumstantial evidence that this weaker version is actually equivalent to R3.
(Only in a computational sense! My hypothesis: If A and B are two diagrams of the same knot, while it might not be actually possible to go from A to B by applying weak R3 moves (+R2+R1, of course), the assumption that weak R3 holds forces invariant(A)=invariant(B) for any Lie group derived invariant. Likewise, in Kauffmans abstract tensor framework, just assume weak R3, solve and get the rest of the YangBaxter equation for free.)
Thus: Is there work on "alternative moves"? Can you construct a counterexample? (I.e. an pseudoinvariant which is constant under weak R3+R2+R1, but not under R3? The example must have a skein equation, though.)

4$\begingroup$ I believe no one has responded/voted to this yet because it comes across incomprehensible. Can you state things more clearly? (Maybe with definitions) $\endgroup$ – Chris Gerig Sep 14 '12 at 5:12

$\begingroup$ Blast, not even the diagram came through. I try a total workover. $\endgroup$ – Hauke Reddmann Sep 14 '12 at 8:42

$\begingroup$ @Hauke: I fixed the image reference. $\endgroup$ – Joseph O'Rourke Sep 14 '12 at 10:00
I am very very skeptical that you can recover R3 from your weak R3. Instead here's what I expect is happening in your examples. Let V be the vector space spanned by diagrams with 6 boundary points. There's a pairing on V given by connect up all the boundary points and apply your invariants. You might as well treat as 0 anything in V that's in that's in the radical of this form, it won't change your invariants. So what we want to know is whether the difference of the two sides of R3 (which I'll call v) is in the radical of this form for your particular examples. When you take inner products of v with various other vectors in V you do the calculation by applying whatever skein relations you have. It's quite possible that in the presence of your other specific skein relations your weak R3 is all you need in order to show that v has inner product 0 with everything. But this would be because you've imposed enough other special skein relations.

$\begingroup$ (@Joseph  THX, I read the manual for image insertion twice and still got it wrong...) @Noah: The other skein relations are not THAT special  I'll try my best to write up a formal proof but I never excelled in that :) My main observation was that I can write the conditions for BiedenharnElliott and R3 for 3nodes with 4 boundary points instead of the usual 5. This must "mean" something :) $\endgroup$ – Hauke Reddmann Sep 15 '12 at 14:37