Not exactly an answer, but an omnibus comment:
Re @Charles' answer: from the work of T. Weigel, it is known that if the surjection is onto for SOME prime greater than three, than the group is Zariski-dense. So, the OP's experiments show that the group IS Zariski-dense. Weigel's paper is a bit hard to find (also a bit hard to read), but it is here.
There are groups (first examples were constructed by Steve Humphries in the 1980s) which are infinite index in $Sp(4, \mathbb{Z}),$ and such that the all the congruence projections are onto.
2a. It is an (unpublished) observation of mine that this property is in fact true with positive probability (that is, if you pick, in some reasonably well defined sense, a pair of matrices in $Sp(2n, \mathbb{Z}),$ the group they generate is almost certainly a free zariski dense, group which is profinitely dense with positive probability.
- There is a presentation of $SP(4, \mathbb{Z})$ with two generators and not so many relations:
BENDER, P. “Presentation of Symplectic Group Sp(4,Z) with 2 Generatrices and 8 Definitive Relations.” Journal of Algebra 65, no. 2 (1980): 328-331.
If you can express your matrices in terms of these generators, you can try GAP's "finitely presented groups" package, which can sometimes tell you the answer.
Your general question is most likely undecidable, see: http://mathoverflow.net/questions/47961/undecidability-in-matrix-groups
If the second matrix is a transvection or even just unipotent (trace is 4, so it might be), then you can use to Chris Hall's theorem Big symplectic or orthogonal monodromy modulo $\ell. $ C Hall - Duke Mathematical Journal, 2008 to check whether it surjects onto any given congruence quotient.
