Let $$R(1) = 1-1,$$ $$R(2) = (1\times11) - (1+11),$$ $$R(3) = (1\times11\times111) - (1+11+111),$$ and so on...
$$R(4)=1355297\quad\text{(a prime number!)}$$
$R(4)$ is the only prime I found of such form up to $R(200)$. Are there anymore primes of such form?
Probabilistic answer: If $\phi_n\in {\Bbb N}$ is a sequence of 'random' integers going to infinity then the probability of $\phi_n$ being a prime is $\sim 1/\ln \phi_n$. When $\sum_{n\geq N} \frac{1}{\ln \phi_n}$ goes to zero fast with $N$ it is most 'likely' that there are no primes among these numbers for $N$ large. The words 'random' and 'likely' are obviously subject to interpretations.
If another prime of the form $R(n)$ exists, then $n$ must be at least $490$ and $R(n)$ must have more than $100,000$ digits.
For the following numbers $n\le 1000$, $R(n)$ has no "small" prime factor :
$$[52, 490, 532, 574, 592, 922, 928, 964]$$
$R(1000)$ has already $499,546$ digits.
It is unlikely that there is another prime $R(n)$ because the sequence grows very fast and the chance that a number with more than $10^5$ digits is prime, is very low. Of course, this is not a proof.
A proof that there is no other prime should be out of reach. The only chance seems to be : Finding another prime.
Just a very basic observation, not a complete answer. Let's note by $$a_n=1111...1$$ with $n$ of $1's$. So we have $$a_n \equiv 1 \pmod{10}$$ And $$\prod_{k=1}^{n} a_k \equiv 1 \pmod{10}$$ And $$\sum_{k=1}^{n} a_k \equiv n \pmod{10}$$ And $$R(n)=\prod_{k=1}^{n} a_k - \sum_{k=1}^{n} a_k \equiv 1 - n \pmod{10}$$ So, whenever $n-1$ is divisible by $2,5$ or $10$, $R(n)$ is not a prime.
And another one is $$a_n \equiv \sum_{k=1}^{n}1=n \pmod{9}$$ So $$\prod_{k=1}^{n} a_k \equiv n! \pmod{9}$$ And $$\sum_{k=1}^{n} a_k \equiv \frac{n(n+1)}{2} \pmod{9}$$ And $$R(n)=\prod_{k=1}^{n} a_k - \sum_{k=1}^{n} a_k \equiv n! - \frac{n(n+1)}{2} \pmod{9}$$ For $n \geq 6$ with $\frac{n(n+1)}{2}$ divisible by 3, $R(n)$ is not a prime. Out of $n=6t$, $n=6t+1$, $n=6t+2$, $n=6t+3$, $n=6t+4$ and $n=6t+5$ only $n=6t+1$ and $6t+4$ could yield primes. Because $n=6t+1$ is clarified by the case 1 ($n-1 = 6t$ divisible by $2$) we are left with $6t+4$.
I wanted to post this as a comment but it was too long.
I don't know if it will help but consider to rewrite your function in this way:
$$R(n) = (1 \times 11 \times 111 \times \ldots) - (1 + 11 + 111 + 1111 + \ldots)$$
$$R(n) = \left((10^0) \times (10^0 + 10^1) \times (10^0 + 10^1 + 10^2) \times \ldots\right) - \left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right)$$
Now with a bit of maths you can check that
$$\left((10^0) \times (10^0 + 10^1) \times (10^0 + 10^1 + 10^2) \times \ldots\right) = \prod_{k = 1}^n \frac{10^k - 1}{9}$$
For what concerns the second part, it's more amusing. Indeed we have
$$\left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right)$$
But as we tend to $n$, we can easily check that there will be $n$ terms of $10^0$, $n-1$ terms of $10^1$, $n-2$ terms of $10^2$ and so on, which means
$$\left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right) = \sum_{k = 0}^{n-1} (n-k)10^k$$
Now with a bit of help (mathematica rules), the productory gives
$$\prod_{k = 1}^n \frac{10^k - 1}{9} = 9^{-k} {Q}_p (10, 10, k)$$
Where ${Q}_p (10, 10, k)$ is the so called q-Pochammer symbol, whose definition is
$${Q}_p (a, q, k) = \prod_{i = 0}^{k-1} (1 - aq^i)$$
Ref https://en.wikipedia.org/wiki/Q-Pochhammer_symbol
Whereas the sum gives
$$\sum_{k = 0}^{n-1} (n-k)10^k = \frac{1}{81}\left(10^{1+n} - 9n - 10\right)$$
Thus your function is
$$R(n) = 9^{-k} {Q}_p (10, 10, k) - \frac{1}{81}\left(10^{1+n} - 9n - 10\right)$$
As I said, I don't know if this helps, but it's a good way to check $R(n)$ quite fast.
Please If I made some mistake, make a comment. I liked this question and I immediately got involved in trying to find a suitable general form for $R(n)$.
