The following answer comprises a collection of remarks.
(1) What, typically, allows a quantum version of a classical group is not relations between elements of the group $G$ (but see (2)), but rather between functions on the group.
For example, where $\mathbf{1}_{j\to i}\in C(S_N)$ is the indicator function for the subset of permutations in $S_N$ that map $j$ to $i$ ($i,j\in\{1,2,\dots,N\}$), we have that these indicator functions are projections in the $\mathrm{C}^*$-algebra $C(S_N)$, and, if we populate a matrix $v\in M_N(C(S_N))$ according to $v_{ij}=\mathbf{1}_{j\to i}$, they also they sum to one along $i=1,\dots,N$ and along $j=1,\dots N$ (note if we are in the $\mathrm{C}^*$-algebraic setting orthogonality for such partitions of unity is automatic (but see here)). Such a matrix is called a magic unitary.
So, the relations between these functions, which generate all of $C(S_N)$, means that the algebra of functions can be given as a universal commutative $\mathrm{C}^*$-algebra: $$C(S_N)=\mathrm{C}_{\operatorname{comm}}(v_{ij}\mid v\text{ a magic unitary}).$$
Now drop commutativity and you get the algebra of continuous functions on the quantum permutation group $S_N^+$. These days the notation $C(S_N^+)$ is used for $A_s(N)$.
This approach allows us to define a quantum orthogonal group, a quantum unitary group, and a quantum automorphism group of a graph. And sitting inside all of these, via the abelianisation, are the classical versions they have "liberated".
Forgetting all about topology, I imagine there are no nice relations on functions on the braid group.
(2) I would say that the typical place that we see generators in the group giving a quantum group is the dual case. So, for example, take whatever presentation you want of $S_N$ and form the group $\mathrm{C}^*$-algebra from this presentation. Then you can give this the structure of the algebra of continuous functions on compact quantum group, the dual of $S_N$: $$C(\widehat{S_N}):=\mathrm{C}^*(S_N).$$
I imagine this can be done for the braid group no problem... but note $\widehat{S_N}$ isn't a quantum version of $S_N$... in fact it is a liberation of $\mathbb{Z}_2$. I think the liberation of $\widehat{B_N}$ should be an abelian group... as a set equal to the set (equivalence classes?) of one dimensional representations of $B_N$.
(3) I have not checked if applicable, but Section 3 of Rollier--Vaes might produce a quantisation of the braid group...
(4) As an addenda, I was just in a seminar talk where it was claimed as a moral that actions of the (possibly infinite) symmetric group in classical probability should be replaced by action of braid groups in quantum probability... see here.