No. Let $k = \mathbb{Q}(\sqrt{-17})$. The class group of $k$ is $\mathbb{Z}/4 \mathbb{Z}$, generated by $\langle 3, 1+\sqrt{-17} \rangle$. (I checked this table for a field of class number $4$ and then checked that $\langle 3, 1+\sqrt{-17} \rangle$ and $\langle 3, 1+\sqrt{-17} \rangle^2$ are not principal. To do this, note that $a^2+17 b^2=3$ is not solvable in integers, and $a^2+17 b^2=9$ only has the solution $(\pm 3, 0)$.)

Let $L$ be the class field of $k$. The only fields between $L$ and $k$ are $L$, $k$ and an intermediate quadratic extension $K$, corresponding to the three subgroups of $\mathbb{Z}/4 \mathbb{Z}$. But $k(\sqrt{-1})$ is abelian and unramified, so $K$ must be $k(\sqrt{-1})$. Thus, all nontrivial extensions of $k$ within $L$ contain a root of unity.

The field $L$ is $k(j(\sqrt{-17}))$, where $j$ is the $j$-function. The minimal polynomial of $j(\sqrt{-17})$ is

```
x^4 - 178211040000 x^3 -75843692160000000 x^2 -318507038720000000000 x -2089297506304000000000000,
```

and an explicit formula in radicals is
$$8000 \left(\ 5569095 + 1350704 \sqrt{17} +
4 \sqrt{2 (1938444620639 + 470141877665 \sqrt{17})}\ \right).$$
So this is clearly a quadratic extension of $\mathbb{Q}(\sqrt{-17}, i)$. It isn't easy to me to see that it is Galois over $\mathbb{Q}(\sqrt{-17})$, though.

I note that the element $2(1938444620639 + 470141877665 \sqrt{17})$ in $\mathbb{Q}(\sqrt{17})$ has norm $-421496^2 = - 2^6 19^2 47^2 59^2$ in $\mathbb{Q}$.

I think the prettiest way to describe the top field is $\mathbb{Q}(\sqrt{1+4i}, \sqrt{1-4i})$. This is clearly Galois over $\mathbb{Q}$ with Galois group dihedral of order $8$. Writing $\rho$ for the order $4$ rotation in this dihedral group, the fixed fields of $\rho$ and $\rho^2$ are $\mathbb{Q}(\sqrt{-17})$ and $\mathbb{Q}(\sqrt{17}, i)$ respectively, and it isn't too bad to check that $\mathbb{Q}(\sqrt{1+4i}, \sqrt{1-4i})/\mathbb{Q}(\sqrt{-17})$ is unramified..