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(Edit: I edited a bit. Now the answer should be clearer, simpler and even correct. Thank you Andrei Smolensky for repeating correcting my embarrassing mistakes here.)

I am surprised that this old question was not fully answered yet. The answer is "No" and it is well known in some circles. In fact, a far more general statement holds:

1. Let $R$ and $S$ be rings (commutative with 1). Then any group homomorphism $\text{SL}_3(R)\to\text{SL}_2(S)$ factors via the non-trivial quotient $\text{SK}_1(3,R):=\text{SL}_3(R)/\text{EL}_3(R)$, where $\text{EL}_3(R)$ is the subgroup generated by elementary matrices in $\text{SL}_3(R)$.

Note that $\text{SL}_3(R)$ contains an epimorphic image of $\text{SL}_3(\mathbb{Z})$ (induced by the map $\mathbb{Z}\to R$). It is well known (and easy) that $\mathrm{EL}_3(\mathbb{Z})\simeq \mathrm{SL}_3(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ is contained in $\mathrm{EL}_3(R)$, and in fact $\mathrm{EL}_3(R)$ is generated by $\text{SL}_3(\mathbb{Z})$ as a normal subgroup (as you can observe by playing with commutation relation of elementary matrices). Thus (1) is equivalent to:

2. Let $S$ be a ring (commutative with 1). Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S)$ is trivial.

We now fix a homomorphism as in statement (2) and assume its image is non-trivial. It is standard that there is a maximal ideal $m\lhd S$ and an integer $k$ such that the image of $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m^k)$ is already non-trivial. Note that the the kernel of $\text{SL}_2(S/m^k)\to\text{SL}_2(S/m)$ is nilpotent and that $\text{SL}_3(\mathbb{Z})$ has no non-trivial nilpotent quotients (every elementary matrix is a commutator, thus it is perfect). It follows that $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m)$ is non-trivial. We are left to prove the following statement:

3. Let $k$ be a field. Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(k)$ is trivial.

(here I had before an argument I liked, but I had to replace it by a simpler one.)

Here is a nice exercise:

4. Let $k$ be a field. Then for any group homomorphism $\text{H}(\mathbb{Z})\to\text{SL}_2(k)$, where $\text{H}(\mathbb{Z})$ is the integral Heisenberg group, the image of the center (=commutator group) of $\text{H}(\mathbb{Z})$ consists of scalar matrices.

Hint: Assume the image of a generator of the center is not a scalar matrix and show that $\text{H}(\mathbb{Z})$ is in the Borel, in which every nilpotent group is abelian (you may assume that $k$ is algebraically closed here).

Remark: Actually, the image of the center of $\text{H}(\mathbb{Z})$ will be trivial unless $\text{char}(k)=2$.

To finish up with (3), observe that every elementary matrix in $\text{SL}_3(\mathbb{Z})$ is the center of a conjugate of $\text{H}(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ consists of scalar matrices. But $\text{SL}_3(\mathbb{Z})$ is perfect, so this image is trivial.