By this discussion on John Baez's Google+ feed, the primes $p$ such that the decimal expansion of $1/p$ repeats with period 2017 are exactly those primes which occur in the prime decomposition of $10^{2017} - 1$ (and for which $10$ isn't equal to $1$ modulo $p$ – this excludes $p = 3$).
The first two prime factors of $10^{2017} - 1$ are $3$ and $129089$. How should I go on to check whether the cofactor, a number with 2012 digits, is prime? It isn't listed on the database maintained by Chris Caldwell when displaying all known primes with 2012 digits, so it's probably not prime.
Let's say $p$ divides $10^{2017}-1$. This means: $$10^{2017} \equiv 1 \pmod{p}$$ Using group theory, we know that $2017$ must be a multiple of the order of $10 \pmod p$. The order of $10 \pmod p$ is the smallest number $n$ such that $10^n \equiv 1 \pmod p$. We know that $n \neq 1$ because if $10^1 \equiv 1 \pmod p$, then $p=3$ or $p=9$. However, the only other factor of $2017$ is $2017$, so the order of $10$ must be $2017$.
We also know that the order of $2017$ must divide $\phi(p)$. For a prime $\phi(p)$ is always $p-1$, so this means $2017$ is a divisor of $p-1$, or: $$p-1 \equiv 0 \pmod{2017}$$ $$p \equiv 1 \pmod{2017}$$ Therefore, other than $3$, any prime factors of $10^{2017}-1$ must be in the form of $2017k+1$. For example, $129089=2017k+1$. Checking for prime factors in the form of $2017k+1$ instead of just all numbers will reduce your search space, but only by a constant factor, so unfortunately, there are still way too many possibilities to check them all even using this strategy.