Yes.

An invariant $f$ of braids $b$ is an invariant of links (obtained as braid closures) if it satisfies the Markov moves:
1. $f(\sigma_i b \sigma_i^{-1}) = f(b)$
2. $f(\sigma_n b) = b$ whenever $b$ contains only the generators $\sigma_1, \dots, \sigma_{n-1}$ (i.e., when $b$ is a braid on strands $1, \dots, n$).

Actually, really only the first one is required. Being an invariant of braids requires the RIII and RII relations, and (1) is just another case of the RII relation. (2) is equivalent to the RI move, which is not a move on framed links. A trace-like operation which satisfies (1) (which is equivalent to cyclicity) but not (2) is sometimes called a Markov trace and gives an invariant of framed links.