"Although it is necessary for $n$ to be prime in order for $R_n$ to be prime" as logic statement This is a statement about Repunits from this paper.
How can I write this as an if/then statement?
Knowledge about Repunits isn't required. The question is basically: what does "although it is necessary for $a$ to be prime in order for $b$ to be prime" mean?      
I think it means: if $b$ is prime then $a$ is prime. 
 A: 
although it is necessary for "$a$ to be prime" in order for "$b$ to be prime".

That is, if $b$ is prime, then we know that $a$ must be prime. 
You are right that the right translation is 
"If $b$ is prime, then $a$ is prime" which corresponds to "If $R_n$ is prime, then $n$ is prime"
Notice that in this case we say that "$a$ is prime" is a necessary condition to "$b$ is prime". We say that "$b$ is prime" is a sufficient condition for "$a$ is prime".
A: The logically equivalent statement to yours might be:

If $R_n$ is prime then $n$ is prime.

If you are interested in proving such statement, consider the following "proving the contrapositive":

$R_n$ can be written as $\frac{10^n-1}{9}$ given that the
  fraction $\frac{10^n-1}{9}$ is always an integer. Now, if $n$ is not prime,
  then $n = ab$ for some prime $a$ and integer $b$. Hence, $R_n = \frac{10^{ab}-1}{9}=\frac{(10^a)^b-1}{9}$. But $(10^a)^b-1 = (10^a - 1) \times \sum_{i=0}^{b-1}(10^a)^{i}$ and $R_n = \sum_{i=0}^{b-1}(10^a)^{i} \times \frac{10^a - 1}{9}$ This implies that $(\sum_{i=0}^{b-1}(10^a)^{i})$
  is a divisor of the Repunit $R_n$. Thereby, if $n$ is not prime, then so is $R_n$. Contrapositively, $n$ must be prime if $R_n$ is prime.

You may read this post for more details about repunits.
