Preface: Please bear my lousy ASCII :-)
I automatically thought that a twist move on a trivalent node -(/ = a* -( would be diagonal even in the case of multiplicity, for example: -1(/= a* -1( + 0* -2(. But thinking twice, of course you now could rotate the basis (1,2) and get cross terms, so a priori the R symbol could have off-diagonal terms.
It's ten seconds work with Reidemeister 3a moves to see that (for real irreps) the R-symbol must be independent of the order of the three irreps i,j,k around the node (up to the twist factor $r_ir_j/r_k$), and matrix-wise fulfil $R^+=R, R*R=I$. But still the question remains: Assuming (natural choice) your basis is Schur-orthogonal, -1()2- = 0. Is the R-symbol necessarily diagonal then?
Here is a quick argument for "no": Take the fusion category F=2148 $(A\bigotimes{A}=E+2A+B,B\bigotimes{B}=E+8B+4A,A\bigotimes{B}=A+4B)$. You can check that F is braided (I omit the twist data due to two ugly nested roots). If R is diagonal, the cross-term projectors A12, A21, read )2-1( , in the Clebsch-Gordan expansion of the braiding tensor S vanish and it "lives" in a 4-dim vector space (E11,A11,A22,B11). Since there is a left and a right A in $A\bigotimes{A}=E+2A+B$ (like the 7 in $7\bigotimes7=1+7+14+27$), this means you can adapt the proof that your quantum group must be $G_2$(something) - the multiplicities don't interfere here. But for any q in $G_2(q)$, you won't get the quantum dimensions $A=2+\sqrt{6}, B=5+2*\sqrt{6}$ correct. Contradiction.
So, could you please a) shoot a hole in my proof - in a Schur-orthogonal basis the R-symbols are diagonal (e.g. it seems to be the case for $SU(3)_3/Z_3$, cf. Ardonne/Slingerland, where R=diag(1,-1) for the 8 in $8\bigotimes8=1+8+8+10+10'$) or b) confirm that I was right, best by citing an example of a fusion rule with multiplicity and known, off-diagonal R symbols.