Yes. I just noticed that if we let $x_{i}=2$ for all $i$, we see that $\mathbb{Q}$ also doesn't work. :-)

Gotcha. I thought that the example with $\mathbb{Z}$ is well known. I've figured out a simplification to my counter-example to your original question. Just let $K=\mathbb{Q}(x_{1},x_{2},\ldots)$. Notice that no element has $2^{n}$ roots for all $n$ (except $1$) by a simple degree argument. Then consider $(x_{1}^{2},x_{2}^{4},x_{3}^{8},\ldots)$ (modulo the appropriate subgroup). etc...

Re: Ryan -- As one can express real numbers such that there is no algorithm to determine if they are equal to zero, the answer is still no. [So, in some ways that makes my answer trivial, since then polynomials work. But see the other comment below.] Re: Mariano -- That's the big question! How much information can we give and still be unable to determine whether such a function has a zero? For a concrete example: Given the L-function associated to a rational elliptic curve of rank 4, can we algorithmically prove the analytic rank is 4? From what I understand this is still open!