Inspired by this post, I established an even stronger conjecture:
If we have the number $1\ldots 1$ with $k\ge 3$ digits $1$ in decimal expansion, then we can replace a digit which is neither the first nor the last by one of $2,3,\ldots ,9$, to get a prime number.
More mathematically formulated
Let $k\ge 3$ be an integer. Then, there exists an integer $j$ with $1\le j\le k-2$ and an integer $m$ with $1\le m\le 8$, such that $$\frac{10^k-1}{9}+m\cdot 10^j$$ is prime (which is a near-rep digit prime number as desired in the question).
This is a significant strengthening of what is asked in the question.
Heuristic: The desired prime is roughly $10^{k-1}$, a random number with approximately this magnitude is prime with probability about $\frac{1}{\ln(10)\cdot (k-1)}$ and since we have the choice of $8\cdot (k-2)$ numbers, for $k\ge 1000$, the expected number of primes is about $3.47$.
Computational result: The conjecture is true upto $k=1000$ (I proved the primality of the found examples until $k=500$ , for the rest of the range the very realiable BPSW-test was used).
Any idea, reference or extension of the search limit would be very welcome.