In this paper they talk about this problem for 5 instead of 10 roots.
http://www.jstor.org/stable/2323469
EDIT: In view of Todd Trimble's comment, here's a summary of what's in the paper.
Let $f(k,N)$ be the least absolute value of a nonzero sum of $k$ (not necessarily distinct) $N$-th roots of unity. Then
$f(2,N)$ is asymptotic to $cN^{-1}$, where $c$ is $2\pi$ for even $N$, $\pi$ for odd $N$,
$f(3,N)$ is asymptotic to $cN^{-1}$, where $c$ is $2\pi\sqrt3$ for $N$ divisible by 3, $2\pi\sqrt3/3$ otherwise,
$f(4,N)$ is asymptotic to $cN^{-2}$, where $c$ is $4\pi^2$ for even $N$, $\pi^2$ for $N$ odd,
$f(k,N)>k^{-N}$ for all $k,N$,
$f(2s,N)<c_sN^{-s}$ for $N$ even and $s\le10$,
$f(k,N)<c_kN^{-[\sqrt{k-6}]-1}$ for $N$ even and $k>5$, and
If $N$ is twice a prime, and $k<N/2$, then there exists $k'<2k$ such that $f(k',N)\le2k2^{k/2}\sqrt{k!}N^{-k/2}$.
The only result in the paper for 5 roots of unity is (the trivial) $f(5,N)>5^{-N}$, but it is suggested that maybe $f(5,N)>cN^{-d}$ for some $d$, $2\le d\le3$, and some $c>0$.
Just as a minor warning: even if the conductor is $1$, there might be nontrivial roots of unity in the class field: take $K = {\mathbb q}(\sqrt{-5}\,)$ and ${\mathfrak c} = (1)$;
then the ray class field is the Hilbert class field $K(\sqrt{-1})$, which contains the 4th roots of unity. The roots of unity in the Hilbert class field (i.e. for conductor $1$) lie in the genus class field and can be computed easily.
Any additional roots of unity must come from ramified extensions; a necessary condition for the $p$-th roots of unity to lie in the ray class field must be that the ry class number, which is easily computed, be divisible by $p-1$ (or $(p-1)/2$ if the genus class field contains the quadratic subfield of the $p$-th roots of unity).
Best Answer
Looks to me like this is false. Let $K = \mathbb{Q}(z)/(z^p-1-p^2)$. This is an extension of degree $p$, so it is disjoint from the p-th cycloctomic field, and hence does not contain a $p$-th root of $1$. Thus, it also can not contain a $p^k$-th root of 1.
Now, let's see how $z^p - 1 - p^2$ factors in Qp. There is already one p-th root of $1+p^2$ in Qp; call this root a. (To see this, note that the power series $(1+x)^{1/p} = 1+(1/p) x + \binom{1/p}{2} x^2 + ...$ converges for $x=p^2$.)
Let $\mathcal{P}$ be a prime of $K$ corresponding to a factor of $z^p-1-p^2$ other than z-a. (In fact, $( z^p-1-p^2)/(z-a)$ is irreducible over $\mathbb{Q}_p$, but I don't need that.) So $K_\mathcal{P}$ contains a root b of $z^p-1-p^2$ other than $a$. But then $b/a$ is in $K_\mathcal{P}$ and is a $p$-th root of 1.
This might be true if you ask $K/\mathbb{Q}$ to be Galois, but I would bet against it.