I think the answer is no. The ultraproduct $U$ is naturally a quotient of ${\mathbb Z}^{\infty}$, the direct product of countably many copies of ${\mathbb Z}$. In the obvious quotient map, the image of the direct sum is zero. Now, it is enough to show that: 

Claim: Any homomorphism $ \phi: {\mathbb Z}^{\infty} \to {\mathbb Z}$ that vanishes on the direct sum is identically zero.

Proof: (I learned this from a book by T.Y.Lam):

For a prime number $p$, let $A_p$ be the set of elements in ${\mathbb Z}^{\infty}$ of the form $(a_0, pa_1, p^2a_2,...)$, i.e. the elements whose $i$-th coordinate is divisible by $p^i$. Any element $x \in A_p$ can be decomposed as 

$x= y+z= (a_0, pa_1, \dots, p^{n-1}a_{n-1}, 0,0, \dots ) + p^n (0,0,..,0, a_n, pa_{n+1},..)$

Now, $y$ is in the direct sum, hence $\phi(y)=0$. Also $\phi(z) \in p^n {\mathbb Z}$, which
implies that $\phi (x) \in \cap_{n=1}^{\infty} p^n {\mathbb Z} = \{ 0 \}$ 

Now, choose two distinct primes $p$ and $q$. Since $\gcd(p^n,q^n)=1$, it is easy to see that
$A_p+A_q= {\mathbb Z}^{\infty}$. This implies that $\phi \equiv 0$.