There is a simple answer here, so someone might as well record it.

Let $n$ be a nonzero integer.  If $n = -1$ or $n$ is a square then there is no odd prime number $p$ such that $n$ is a primitive root modulo $p$.  There are no other obvious obstructions.  (It is worth thinking for a second why we do not have to rule out $n$ being a cube, for instance: this is a nice exercise in cyclic group theory.)

There is a famous conjecture that these obvious necessary conditions are the only ones: namely [Artin's Primitive Root Conjecture][1] asserts that for any integer $n$ which is not $0$, $-1$ and not a square, there are infinitely many prime numbers $p$ such that $n$ is a primitive root modulo $p$.  In fact the conjecture is more precise than this: the set of primes $p$ for which such an $n$ is a primitive root is conjectured to have positive relative density among all primes and, at least under some mild additional restrictions, this density is conjectured to be a certain specific number which is independent of $n$:

$ C = \prod_{p \text{ prime}} \left(1- \frac{1}{p(p-1)} \right)$;

this $C$ is known as **Artin's constant**.  This conjecture was proved by C. Hooley in 1967 assuming the Generalized Riemann Hypothesis.  More recently unconditional results have been given by Gupta, R. Murty and Heath-Brown which consider several numbers $n$ at a time and show that Artin's Conjecture must be true for at least one of them.  But the conjecture is still open for any one fixed value of $n$.

[1]: http://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots