The question is in the title, and I do not really have anything to add. Nevertheless I had to write something here in order to be able to ask the question. Thanks.
Of course. Take a quadratic nonresidue $1\leq n\leq p1$, then some prime divisor $\ell$ of $n$ will be a quadratic nonresidue. See this MO question for what is known about number fields. 


It is actually quite easy to prove that if $p>3$, then there are at least $2$ primes less than $p$ which are quadratic nonresidues. Indeed, assume there were only one, say $q$. Then every $n$ between $1$ and $p1$ which is not multiple of $q$ is a quadratic residue. Since you have at most $(p1)/q$ multiples of $q$, and exactly $(p1)/2$ quadratic residues, this implies $q=2$ and moreover $p=3$ (since otherwise you would get too many quadratic residues: every odd number between $1$ and $p1$, together with $4$). 


Slightly different in emphasis, the smallest quadratic nonresidue is in fact prime, as the product of residues is another residue. 


Erdos conjectured that for any sufficiently large prime $p$ there is a primitive root $q<p$ for $p$ which is prime. 


I think the answer is obvious. Since $$\sum_{1\leq n\leq p1}\left(\frac{n}{p}\right)=0$$, there must exist a positive integer $n\leq p1$, such that $(\frac{n}{p})=1$, or else the summation above must be equal to $p1$. Of course, maybe $n$ is not a prime, however there always be a prime factor $\ell$ of $n$ such that $(\frac{\ell}{p})=1$. 

