$2k=1+p_N$ works for $N>1$, but $2k\le 0.56 \, p_N$ will fail if $p_{N+2}=p_{N+1}+2$.
With $q=p_{N+1}$, we have $$ \frac{1}{1-q^{-2k}} < \frac{1}{1-x^{-2k}} = \frac{1}{1-q^{-2k}} \prod_{p>q} \frac{1}{1-p^{-2k}} . $$ It follows that $$ q^{-2k} < x^{-2k} < q^{-2k} + \sum_{j\ge 2} (q+j)^{-2k} < q^{-2k} +\frac{1}{(q+1)^{2k-1}(2k-1)}. $$ Taking logarithms, and using $\log(1+y)\le y$, we get $$ -2k \log q < -2k \log x < -2k\log q + \frac{q+1}{\exp\{(1-o(1)) 2k/q\} (2k-1)}. $$ Dividing by $-2k$ and exponentiating, we have $$ q > x > q - \frac{q(q+1)}{\exp\{(1-o(1))2k/q\} 2k (2k-1)}. $$ We want the last expression to be less than $1/2$. Since $q/p_N \to 1 $ as $N\to \infty$, we need $k\ge (1+o(1)) c \, p_N$, where $c=0.45...$ is the solution to $e^{2c}4 c^2=2$. So $2k=1+p_N$ works for large $N$. We can check with a computer that it also works for small $N$. Similar calculations show that when $p_{N+2}=p_{N+1}+2$, it is necessary that $k>0.28 \, p_N$ when $N$ is large.
Edit: Using the inequality $\log(1+y)\le y$ was somewhat wasteful and not necessary. Also, as the OP points out in the comments, we can use the ceiling function instead of rounding, since $x<q$. With those two modifications, we find that $k\ge \frac{1}{3}p_N$ works for $N\ge 1$, but $k\le 0.19 \, p_N$ fails at twin primes $p_{N+2}=p_{N+1}+2$.