Here is a partial answer. Finite subgroups of SO(4) which act freely on $S^3$ (thought of as the unit sphere in $R^4$) are in (almost) 1-1 correspondence with spherical 3-manifolds. The only discrepancy is with lens spaces $L(p,q) \cong S^3/(Z/pZ)$, which will be discussed at the end of this answer. 

For manifolds with non-cyclic fundamental group, the groups are well understood. As noted in the question, these groups (the ones that act freely) are part of the  classification of Conway and Smith. Because you are asking for a specific representation, a natural perspective is to consider an action on $S^3$. Jeff Weeks' [Curved Spaces][1] program does these computations (although there might be other places they appear, you get to fly around in these spherical spaces, which is really cool. It's perhaps easiest to get the files I am referring to by downloading the source code) To get numerical approximations of the matrices you are looking for, download the source code and look in "Curved Spaces/Source-Common/Assets/Sample Spaces/Spherical" the files *.gen basically are just a list of generators of your groups. For more background the geometry of these files, Thurston's book "Three-Dimneionsal Geometry and Topology" is a natural reference (especially Chapter 4.4). From the perspective of 3-manifold topology, one nice observation that arises is that all finite subgroups of $SO(4)$ acting freely on  $S^3$ have presentations with two generators (this follows from the fact that all elliptic 3-manifolds have genus 2 Heegaard splittings). 
Here is a sample file for the finite subgroup of $SO(4)$ acting freely on $S^3$ of order 120, aka $\pi_1$ of the Poincaré homology sphere.


(Note: the entries have been clipped to fit this formatting. For the full file see
"Curved Spaces/Source-Common/Assets/Sample Spaces/Spherical/Binary Icosahedral\ L.gen")
> #      Binary Icosahedral Group
>#
>#       The quotient of the 3-sphere by the binary icosahedral group
>#       is the famous Poincaré dodecahedral space.
>#       The matrix elements lie in the field Q[φ], where
>#       φ is the golden ratio.  The golden ratio is a root
>#       of the irreducible polynomial φ² - φ - 1,
>#       with numerical value φ = (1 + √5)/2 ≈ 1.6180339887...
>
>  0.809016994374947 -0.500000000000000  0.000000000000000 -0.30901699437494742410
>  0.500000000000000  0.809016994374947  0.309016994374947  0.000000000000000
>  0.00000000000000 -0.309016994374947  0.809016994374947  0.500000000000000
>  0.309016994374947  0.000000000000000 -0.500000000000000  0.80901699437494
>
>  0.809016994374947 -0.309016994374947  0.500000000000000  0.00000000000000
>  0.30901699437494  0.80901699437494  0.00000000000000  0.50000000000000000000
> -0.5000000000000  0.00000000000000  0.809016994374947  0.309016994374947
>  0.000000000000000 -0.50000000000000 -0.309016994374947  0.809016994374947


(To get an exact representation use $0.809016994374947\approx (1+\sqrt{5})/4, -0.309016994374947 \approx (1-\sqrt{5})/4$


 


Finally, let's wrap up with lens spaces. To generate the quotient of $S^3$ by a cyclic group are determined by rotations of order $p$ acting on $R^4$ thought of as $R^2\times R^2$ for the lens space $L(p,q)$ with $p,q$ relatively prime a natural representation of its fundamental group is generated by:

$g = \pmatrix{ \cos(2\pi/p) & \sin(2\pi/p) & 0 &0 \\
-\sin(2\pi/p) & \cos(2\pi/p) & 0 &0 \\
0 & 0 &\cos(2q\pi/p) & \sin(2q\pi/p)  \\
 0 &0 &-\sin(2q\pi/p) & \cos(2q\pi/p)  \\
}$
 
The fundamental groups of lens spaces with equal values of $p$ are isomorphic however, the choice of $q$ affects the homeomorphism type of the manifold (and hence accounts for the noise in the 1-1 correspondence referenced above.) 



  [1]: http://geometrygames.org/CurvedSpaces/index.html