Yes, but you need to check a normalization condition. (Thanks to Student for noting the error in my previous answer.)

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. $f(\sigma_n b) = f(b)$ whenever $b$ contains only the generators $\sigma_1, \dots, \sigma_{n-1}$ (i.e., when $b$ is a braid on strands $1, \dots, n$).

The first condition is essentially the same as braid isotopy: I think you can always use an ambient isotopy to reduce it to applying either the invertibility of the braid generators or a braid-type relation. However, the second condition is a bit more subtle: you need something like it to deal with "extra" strands in your braid, but it is not the same as braid istopy and also fails for framed links, which are usually the natural objects to consider in this context.

One way to deal with this is to replace 2 with a weaker condition:

2'. There is a constant $x$ so that $f(\sigma_n b) = x f(b)$ whenever $b$ contains only the generators $\sigma_1, \dots, \sigma_{n-1}$ (i.e., when $b$ is a braid on strands $1, \dots, n$).

This will give you an invariant of framed links that changes by a power of $x$ when you change the framing, which is typically what happens for quantum invariants.

Ko, Ki Hyoung; Smolinsky, Lawrence, The framed braid group and 3-manifolds, Proc. Am. Math. Soc. 115, No. 2, 541-551 (1992). ZBL0760.57007.

Yes, but you need to check a normalization condition. (Thanks to Student for noting the error in my previous answer.)

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. $f(\sigma_n b) = f(b)$ whenever $b$ contains only the generators $\sigma_1, \dots, \sigma_{n-1}$ (i.e., when $b$ is a braid on strands $1, \dots, n$).

The first condition is essentially the same as braid isotopy: I think you can always use an ambient isotopy to reduce it to applying either the invertibility of the braid generators or a braid-type relation. However, the second condition is a bit more subtle: you need something like it to deal with "extra" strands in your braid, but it is not the same as braid istopy and also fails for framed links, which are usually the natural objects to consider in this context.

One way to deal with this is to replace 2 with a weaker condition:

2'. There is a constant $x$ so that $f(\sigma_n b) = x f(b)$ whenever $b$ contains only the generators $\sigma_1, \dots, \sigma_{n-1}$ (i.e., when $b$ is a braid on strands $1, \dots, n$).

This will give you an invariant of framed links that changes by a power of $x$ when you change the framing, which is typically what happens for quantum invariants.

For a more systematic answer you can keep track of the framings as part of the braid instead of using the blackboard framing; this leads to the framed braid groups discussed in [1].

[1] Ko, Ki Hyoung; Smolinsky, Lawrence, The framed braid group and 3-manifolds, Proc. Am. Math. Soc. 115, No. 2, 541-551 (1992). ZBL0760.57007.

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.

Yes, but you need to check a normalization condition. (Thanks to Student for noting the error in my previous answer.)

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. $f(\sigma_n b) = f(b)$ whenever $b$ contains only the generators $\sigma_1, \dots, \sigma_{n-1}$ (i.e., when $b$ is a braid on strands $1, \dots, n$).

The first condition is essentially the same as braid isotopy: I think you can always use an ambient isotopy to reduce it to applying either the invertibility of the braid generators or a braid-type relation. However, the second condition is a bit more subtle: you need something like it to deal with "extra" strands in your braid, but it is not the same as braid istopy and also fails for framed links, which are usually the natural objects to consider in this context.

One way to deal with this is to replace 2 with a weaker condition:

2'. There is a constant $x$ so that $f(\sigma_n b) = x f(b)$ whenever $b$ contains only the generators $\sigma_1, \dots, \sigma_{n-1}$ (i.e., when $b$ is a braid on strands $1, \dots, n$).

This will give you an invariant of framed links that changes by a power of $x$ when you change the framing, which is typically what happens for quantum invariants.

Ko, Ki Hyoung; Smolinsky, Lawrence, The framed braid group and 3-manifolds, Proc. Am. Math. Soc. 115, No. 2, 541-551 (1992). ZBL0760.57007.