Bruce,
I can give you representants for an open piece of the moduli space (3,3) resp. (3,2,1). I'll add them in Mathematica-form so that you can plug them into any braid you like to work on. The variables a,b,d,x,y,z are the coordinates of the moduli space and l stands for a third root of unity (i dont know how to tell Mathematica to work with roots of unity, so if you know replace l by that thing).
I obtained these representants from the hexagon of 1-dml simples of the modular group and taking the dimension vector (1,1,1,2,1,0) as required. The corresponding moduli problem is then rational (in x,y,z) over that of pairs of 2x2 matrices of rank one. these are rational (in a,b,d). Hope this helps (and sorry for the lengthy formulas below).

sigma1={{-(1 - a - b - d + a d + x - a x - b x - d x + a d x + l x - a l x - b l x - d l x + a d l x - y + a y) z, -(-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, -(-1 + l) z, -(-1 + l) (1 - a - b - d + a d + l - a l - b l - d l + a d l - y + a y) z, (-1 + l) z, (b + d - a d) (-1 + l) z}, {-(-1 + a) (-1 + l) (1 + l) x y z, -(-b - d + a d + l - a l + b x + d x - a d x - l x + a l x + y - a y + l y - a l y) z, (-1 + l) (-1 + x) z, (-1 + a) (-1 + l) (l + x) y z, -(-1 + l) (-1 + x) z, -(b + d - a d) (-1 + l) (-1 + x) z}, {b (-1 + l) (1 + l) x y z, b (-1 + l) (-1 + x + y + l y) z, (a + b - a d - l + d l - a x - b x + a d x + l x - d l x - a y + l y) z, -b (-1 + l) (l + x) y z, (-1 + l) (a + b - a d - a x - b x + a d x - a y) z, -b (-1 + l) (-1 + x + y) z}, {-(1 - a - b - d + a d) (-1 + l) (1 + l) x z, (-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, (-1 + l) z, -(1 - a - b - d + a d + l - a l - b l - d l + a d l + x - a x - b x - d x + a d x + l y - a l y) z, -(-1 + l) z, -(b + d - a d) (-1 + l) z}, {-b (-1 + l) (1 + l) x y z, - b (-1 + l) (-1 + x + y + l y) z, (-1 + l) (1 - d - x + d x - y) z, b (-1 + l) (l + x) y z, -(-1 + d + a l + b l - a d l + x - d x - a l x - b l x + a d l x + y - a l y) z, b (-1 + l) (-1 + x + y) z}, {(-1 + a) (-1 + l) (1 + l) x y z, (-1 + a) (-1 + l) (-1 + x + y + l y) z, -(-1 + l) (-1 + x) z, -(-1 + a) (-1 + l) (l + x) y z, (-1 + l) (-1 + x) z, -(-1 + a + b l + d l - a d l + x - a x - b l x - d l x + a d l x + y - a y) z}}

sigma2={{-(1 - a - b - d + a d + x - a x - b x - d x + a d x + l x - a l x - b l x - d l x + a d l x - y + a y) z, -(-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, -(-1 + l) z, (-1 + l) (1 - a - b - d + a d + l - a l - b l - d l + a d l - y + a y) z, -(-1 + l) z, -(b + d - a d) (-1 + l) z}, {-(-1 + a) (-1 + l) (1 + l) x y z, -(-b - d + a d + l - a l + b x + d x - a d x - l x + a l x + y - a y + l y - a l y) z, (-1 + l) (-1 + x) z, -(-1 + a) (-1 + l) (l + x) y z, (-1 + l) (-1 + x) z, (b + d - a d) (-1 + l) (-1 + x) z}, {b (-1 + l) (1 + l) x y z, b (-1 + l) (-1 + x + y + l y) z, ( a + b - a d - l + d l - a x - b x + a d x + l x - d l x - a y + l y) z, b (-1 + l) (l + x) y z, -(-1 + l) (a + b - a d - a x - b x + a d x - a y) z, b (-1 + l) (-1 + x + y) z}, {(1 - a - b - d + a d) (-1 + l) (1 + l) x z, -(-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, -(-1 + l) z, -(1 - a - b - d + a d + l - a l - b l - d l + a d l + x - a x - b x - d x + a d x + l y - a l y) z, -(-1 + l) z, -(b + d - a d) (-1 + l) z}, {b (-1 + l) (1 + l) x y z, b (-1 + l) (-1 + x + y + l y) z, -(-1 + l) (1 - d - x + d x - y) z, b (-1 + l) (l + x) y z, -(-1 + d + a l + b l - a d l + x - d x - a l x - b l x + a d l x + y - a l y) z, b (-1 + l) (-1 + x + y) z}, {-(-1 + a) (-1 + l) (1 + l) x y z, -(-1 + a) (-1 + l) (-1 + x + y + l y) z, (-1 + l) (-1 + x) z, -(-1 + a) (-1 + l) ( l + x) y z, (-1 + l) (-1 + x) z, -(-1 + a + b l + d l - a d l + x - a x - b l x - d l x + a d l x + y - a y) z}}

EDIT March 9th : The component (3,3;3,2,1) is not able to detect braid-reversion. On the other hand, the component of 6-dimensional representations (3,3;2,2,2) can. A minimal braid that can be separated by this family from its reversed braid is s1^-1s2^2s1^-1s2. One can show that every irreducible component of simple B(3)-representations has a Zariski dense family parametrized by a minimal rational variety. For representations of dimension <= 11 explicit such families are given in this note.

Each of these families can then be turned into a family of 3-string braid invariants over Q(rho) for rho a primitive 3-rd root of unity by specializing the parameters to random integers. For the family (3,3;2,2,2) mentioned above this can be done in SAGE as follows :

K.=NumberField(x^2+x+1)
a=randint(1,1000)
b=randint(1,1000)
c=randint(1,1000)
d=randint(1,1000)
e=randint(1,1000)
f=randint(1,1000)
g=randint(1,1000)
h=randint(1,1000)
B=matrix(K,[[1,0,0,a,0,f],[0,1,1,0,1,0],[1,1,0,1,0,0],[0,0,1,0,d,e],[0,1,0,b,c,0],[g,0,1,0,0,1]])
Binv=B.inverse()
mat2=matrix(K,[[1,0,0,0,0,0],[0,1,0,0,0,0],[0,0,1,0,0,0],[0,0,0,-1,0,0],[0,0,0,0,-1,0],[0,0,0,0,0,-1]])
mat3=matrix(K,[[1,0,0,0,0,0],[0,1,0,0,0,0],[0,0,l^2,0,0,0],[0,0,0,l^2,0,0],[0,0,0,0,l,0],[0,0,0,0,0,l]])
s1=h*Binv*mat3*B*mat2
s2=h*mat2*Binv*mat3*B
s1inv=s1.inverse()
s2inv=s2.inverse()

One can then check braid-reversion for all knots with at most 8 crossings that are closures of 3-string braids. Here are the tests to perform

test41=(s1inv*s2*s1inv*s2-s2*s1inv*s2*s1inv).trace()
test52=(s1inv**3*s2inv*s1*s2inv-s2inv*s1*s2inv*s1inv**3).trace()
test62=(s1inv**3*s2*s1inv*s2-s2*s1inv*s2*s1inv**3).trace()
test63=(s1inv**2*s2*s1inv*s2**2-s2**2*s1inv*s2*s1inv**2).trace()
test73=(s1**5*s2*s1inv*s2-s2*s1inv*s2*s1**5).trace()
test75=(s1inv**4*s2inv*s1*s2inv**2-s2inv**2*s1*s2inv*s1inv**4).trace()
test82=(s1inv**5*s2*s1inv*s2-s2*s1inv*s2*s1inv**5).trace()
test85=(s1**3*s2inv*s1**3*s2inv-s2inv*s1**3*s2inv*s1**3).trace()
test87=(s1**4*s2inv*s1*s2inv**2-s2inv**2*s1*s2inv*s1**4).trace()
test89=(s1inv**3*s2*s1inv*s2**3-s2**3*s1inv*s2*s1inv**3).trace()
test810=(s1**3*s2inv*s1**2*s2inv**2-s2inv**2*s1**2*s2inv*s1**3).trace()
test816=(s1inv**2*s2*s1inv**2*s2*s1inv*s2-s2*s1inv*s2*s1inv**2*s2*s1inv**2).trace()
test817=(s1inv**2*s2*s1inv*s2*s1inv*s2**2-s2**2*s1inv*s2*s1inv*s2*s1inv**2).trace()
test818=(s1inv*s2*s1inv*s2*s1inv*s2*s1inv*s2-s2*s1inv*s2*s1inv*s2*s1inv*s2*s1inv).trace()
test819=(s1**3*s2*s1**3*s2-s2*s1**3*s2*s1**3).trace()
test820=(s1**3*s2inv*s1inv**3*s2inv-s2inv*s1inv**3*s2inv*s1**3).trace()
test821=(s1inv**3*s2inv*s1**2*s2inv**2-s2inv**2*s1**2*s2inv*s1inv**3).trace()

The following tests should give non-zero invariants : 6.3,7.5,8.7,8.9,8.10 (which are known as 'flypes' in the theory) and 8.17 (the non-invertible knot with minimal number of crossings). I tried to include all details in the note Rationality and dense families of B(3) representations. All comments to this text are welcome. I like to thank Bruce Westbury for providing me with feedback.