It depends on how large the image $U$ of the global units $\mathcal{O}_K^\times$ in the group $$\prod_\sigma (\mathcal{O}_K/(\sigma(q)))^\times$$
happens to be. It's almost never the full group (because the unit rank is always smaller than the field degree), and in general it will be very far from it.

In the rare case when $U$ is all of that, you can multiply $q$ by a unit to move its images in the other factors (i.e. $\sigma \neq \mathrm{id}$) anywhere you want without changing the prime ideal it generates. But usually $U$ will be much smaller, even when $\mathcal{O}_K^\times$ happens to surject onto each factor $(\mathcal{O}_K/(\sigma(q)))^\times,$ and certainly when it doesn't. For an example, take $K=\mathbb{Q}(\zeta_4)$ and $p=17$: you only have the fourth roots of unity available to adjust $q$. The image of $q=4+\zeta_4$ in the other residue class field $\mathcal{O}_K/(4-\zeta_4)$ is $8$, and you can move it to any of $-2, -8, 2$; but these are all quadratic residues.

All this is not yet a complete answer - it also depends on how, as $p$ varies, the subgroups of $\mathbb{F}_p^\times$ fit into the additive structure of $\mathbb{F}_p$. But in general I expect you'll find that few $p$ will work.

**Edit:** I was being sloppy. In what ways we can move the images of $q$ in the other residue class fields actually depends on the image $U'$ of $\mathcal{O}_K^\times$ in the group $$\prod_{\sigma\neq\mathrm{id}} (\mathcal{O}_K/(\sigma(q)))^\times$$ (one factor less), and this isn't automatically so small in the totally real case where we have $\mathrm{deg}(K/\mathbb{Q})-1$ independent fundamental units, as many as there are factors in the latter product. I can't complete an argument offhand but now I'm inclined to think that infinitely many $p$ might be viable in this case.

**Edit 2014-05-08:** Here is a partial positive result. Assume that

- $K$ has class number $1$,
- $K$ has infinitely many units,
- $K$ is abelian over $\mathbb{Q}$,
- and a suitable set of generalized Riemann hypotheses.

Then there are infinitely many $q$ (and thus $p$) for which you can fully control what happens in at least *one* factor $\mathcal{O}_K/(\sigma(q))$ ($\sigma\neq\mathrm{id}$).

P. J. Weinberger proved in 1973 (*On Euclidean Rings of Algebraic Integers*) that $\mathcal{O}_K$ is Euclidean in this case, not necessarily for the norm. (This doesn't depend on $K$ being abelian.) A key step in the proof is the existence of sufficiently many prime ideals $\mathfrak{q}$ such that a given fundamental unit $\varepsilon\in\mathcal{O}_K^\times$ maps to a primitive root mod $\mathfrak{q}$. (Being fundamental means $\varepsilon$ is not a proper power of a unit and, in particular, not a root of unity.) Somewhat more precisely, his Theorem 4 yields infinitely many such $\mathfrak{q}$ in a given (ray) ideal class; combining this with our assumption 3 we can get infinitely many such $\mathfrak{q}=(q)$ generated by an element congruent to $1$ modulo the conductor and thus fully split. (This may be overkill.)

Now if the image of $\varepsilon$ generates all of $(\mathcal{O}_K/(q))^\times$, then so does $\sigma(\varepsilon)$ in $(\mathcal{O}_K/(\sigma(q)))^\times$ for each $\sigma$, since the image is the same residue class in $\mathbb{Z}/p\mathbb{Z}$. Take your favorite $\sigma\neq\mathrm{id}$ and multiply $q$ by a suitable power of $\sigma(\varepsilon)$ to move the image of this product mod $(\sigma(q))$ to any desired prime residue class, QED.

But it is not clear whether you can then continue to do this in additional residue class fields without destroying what you already have. The first obstacle is that the $\sigma(\varepsilon)$ may not be multiplicately independent. (They certainly won't be unless $K$ is totally real. Even then, $\varepsilon$ might have norm $\pm1$ in a subfield strictly between $K$ and $\mathbb{Q}$.) The second and more serious obstacle is that any further tweaking of $q$ must be done by units with image $1$ in each of the residue class fields we've already dealt with, and we may run out of units mapping to primitive roots before we've dealt with the whole lot.