I have not seen this written down but this can certainly be done explicitly. The construction assumes you have explicit matrices for the representations of the symmetric group. Here you have a choice between Young's orthogonal form (which has rational entries) and Specht modules (which have integer entries).

Once you have sorted this out you can construct explicit matrices for the representations of BMW using a Morita equivalence approach. The irreducible representations of the $r$ string algebra are indexed by partitions $\lambda$ with $r-|\lambda|\ge 0$ and even. Let $V(\lambda)$ be the representation of the $|\lambda|$-string symmetric group. let $M$ be the vector space of Brauer diagrams with $r$ points on the top, $|\lambda|$ points on the bottom, with precisely $\lambda$ through strings and such that no two through strings intersect.
Then the $r$-string BMW algebra acts on $V(\lambda)\otimes M$.

More detail:

Here is the theory behind the construction. First I realised I am thinking of the Brauer category whereas OP asks about BMW. This basically means the symmetric group algebra should be replaced by the Hecke algebra.

The Brauer category has an increasing sequence of ideals. The ideal $I(k)$ is spanned by diagrams with at most $k$ through strings. The quotient $I(k)/I(k-1)$ is spanned by diagrams with precisely $k$ through strings. Then the key idea is that the category $I(k)/I(k-1)$ is Morita equivalent to the group algebra of the $k$-string symmetric group.
The functor from representations of the $k$-string symmetric group to representations of the $k+2c$ string Brauer algebra has a straightforward explicit description.

Even more detail:

The Morita equivalence I have in mind is an extension of the usual equivalence. Let $A$ be a finite dimensional unital algebra and $e\in A$ an idempotent. Let $J$ be the ideal $AeA$
and $B$ the unital algebra $eAe$. Then $J$ and $B$ are Morita equivalent since we have the bimodules $eA$ and $Ae$. Here $J$ is not unital but does have the weaker property that $JJ=J$. Then a right $J$-module $M$ is required to satisfy $MJ=M$. Something like this but without mentioning Morita theory is in Sandy Green's SLN text on Schur algebras.