The question is slightly mistated, from the example for $n=2$ it is seen that the question is about $SL_q(n)$ and not $SU_q(n)$; the answer is of course those standard normally ordered monomials
$(t^1_1)^{a_{11}}(t^1_2)^{a_{12}}...(t^n_n)^{a_{nn}}$
satisfying the condition that at least one of the diagonal exponents $a_{ii}$ is zero. Unlike in $O_q(M_n)$ literal application of the Bergman's diamond lemma does not produce the algorithm, because the diagonal enetries are not one next to another so if one wants to exclude the diagonal extra occurences one needs to go against the semigroup law. This is possible to do with great effort, I have checked this in 1999 with lots of algorithmic combinatorics; namely the set of reductions used is infinite and given algorithmically rather than by explciit explicit formulas. Unlike the general rule advised by Bergman, it is not wise in the straight diamond lemma approach to exclude the nested ambiguities. Some other Grober arguments not relying on standard diamond lemma can give easy answer though.
For generic $q$ it is of course enough to use the classical commutative case and deformation arguments (Edit: alluded in David's answer).
It is not true, what is stated above in the accepted answer that the simple technique for $O_q(M_n)$ via diamond lemma and with the relations taken as reductions works when setting $det_q =1$. Imagine you have expression $(x^1_1)^2 (x^2_2)^2 (x^3_3)^2$ in $SL_q(3)$. How will you use centrality of the quantum determinant to translate this into something what does not have all three diagonal generators ? You need first to rearrange thing to be able to complete to a quantum determinant to exclude a bad diagonal generator, but this is not very compatible with the ordering. It can be done systematically but by now means is trivial or implied by Klimyk-Schmuedgen book.
