I think the answer is no. The ultraproduct $U$ is a quotient of ${\mathbb Z}^{\infty}$ the direct product of countably many copies of ${\mathbb Z}$ in which the direct sum is mapped to zero. Now, it is enough to show that
claim: any homomorphism from ${\mathbb Z}^{\infty}$ to ${\mathbb Z}$ vanishes on the direct sum is identically zero.
Here is a proof for this (I learned it from a book by Lam):
Since any element of the form $x= (a_0, 2a_1, 4a_2,...)$ can be expressed as $x' + 2^n (0,0,..,0, a_n, 2a_{n+1},..)$ with $x'$ belonging to the direct sum, any such homomorphism sends such $x$ to zero. The same argument works if you use powers of $3$. Now, since $gcd(2^n,3^n)=1$, any element in the direct product is a sum of two elements of these forms.
