Suppose I am given a machine that gives me the coefficients $a_1$, $a_2$, $a_3$, ... of a Dirichlet series $$\sum_1^{\infty} \frac{a_n}{n^s} $$ and assume that I know that this Dirichlet series is the Dedekind zeta function of a quadratic number field. Is there any kind of algorithm which allows me to determine whether the number field is real or imaginary?

Not without an upper bound on the absolute value of the discriminant $\Delta$, because any finite list of $a_n$ amounts to a congruence condition on $\Delta$ that is satisfied by infinitely many $\Delta$ of either sign. 


Asking in terms of $B$ how many $a_n$ are needed is equivalent to asking the following question: What is the largest $N$ such that there exists two quadratic characters $\chi_1, \chi_2$ of conductor $<B$ with $\chi_1(n)=\chi_2(n)$ for $n<N$ but $\chi_1(1)\neq \chi_2(1)$? An obvious approach is to note that then $\chi= \chi_1 \chi_2^{1}$ is a character of conductor $<B^2$ with $\chi(n)=1$ for $n<m$, which by the bound for the least quadratic nonresidue problem can only happen for $n< \left(B^2\right)^{1/4\sqrt{e}+o(1)}=B^{1/2\sqrt{e}+o(1)}$ Of course any improvement on this problem would also represent improvement on the least quadratic nonresidue problem, at least for residues modulo primes congruent to $3$ modulo $4$. Given this many you could perform the algorithm of enumerating all the characters of conductor $<B$ and seeing which agree with your sequence. The only efficient algorithm I see requires $B$ coefficients  you simply look to see for which primes $\chi(p)$ is $0$ (or $a_p$ is $0$ for the Dirichlet $L$function). These are precisely the ramified primes. If you check up to $B$ you find all the ramified primes. Knowing the ramified primes determines the Dirichlet character up to multiplication by a Dirichlet character modulo $8$, since $2$ is the only prime that can be ramified in multiple different ways. Simply check the four possibilities to see which one matches the first few coefficients of your sequence. 

