edit, Monday, December 5: Pedja's argument can me made to work with the addition of a quadratic reciprocity step...
original: Pedja, you use Dirichlet on the arithmetic progression $12 n + 7,$ which is what you are describing by saying $6k+1$ with $k$ odd. Write $k = 2n+1,$ then $6k+1 = 6 (2n+1) + 1 = 12n + 6 + 1 = 12 n + 7.$
Meanwhile, if you are willing to accept the Cebotarev Density Theorem, asymptotically the two positive quadratic forms $x^2 + 12 y^2$ and $3 x^2 + 4 y^2$ represent the same proportion of primes. The first form represents primes $p \equiv 1 \pmod {12},$ while the second form represents 3 and the primes $q \equiv 7 \pmod {12}.$ I am quoting Theorem 9.12 on page 188 of Primes of the Form $x^2 + n y^2$ by David A. Cox.
EDIT: Actually, your attempt is almost correct, there is an elementary proof of this one. As you write above, take the collection of primes $q_j \equiv 7 \pmod{12}$ that is assumed to be complete, up to some largest we will call $q_r.$ Then, just as you wrote, define
$$ w = 3 + 4 \left( q_1 q_2 \ldots q_r \right)^2 $$
We know that $w \equiv 7 \pmod {12}.$ We also know that $w$ is primitively represented by the quadratic form $3 x^2 + 4 y^2.$ That means that $\gcd(x,y)=1,$ where here $x=1,$ and we are writing $w = 3 x^2 + 4 y^2.$
Now, $w$ is not divisible by 2 or 3. The fact that $\gcd(x,y)=1$ means precisely that $w$ is not divisible by any prime $s \equiv 2 \pmod 3.$ This fact comes under the general heading of quadratic reciprocity. All prime factors of $w$ are, in fact , $1 \pmod 3.$ That is, all prime factors of $w$ are either $1 \pmod {12}$ or $7 \pmod {12}.$ If all prime factors of $w$ were $1 \pmod {12},$ then $w$ itself would also be $1 \pmod {12},$ which it is not.
So, in fact, either $w$ is prime, or $w$ has at least one prime factor, call it $q_{r+1},$ such that $q_{r+1} \equiv 7 \pmod {12}.$ By construction, for any $j \leq r,$ we have $\gcd(q_j,q_{r+1} ) = 1.$ That is, either $w$ itself or $q_{r+1}$ is not in the assumed finite list of primes $7 \pmod {12},$ contradicting the assumption that there is a finite list. $\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$
EDIT TOO: Maybe not everyone will know this, so I will give the short proof. Suppose we have some prime $s \equiv 2 \pmod 3,$ with $s \neq 2,$ that divides some $ z = 3 u^2 + 4 v^2.$ Then suppose that $s | z,$ which is also written $ z \equiv 0 \pmod s.$ I claim that, in fact, both $s | u$ and $s | v,$ with the result that $\gcd(u,v)$ is larger than one. PROOF: We have $ 3 u^2 + 4 v^2 \equiv 0 \pmod s.$ Assume that $v \neq 0 \pmod s.$ Then $v$ has a multiplicative inverse $\pmod s.$ So we get
$$ 3 u^2 + 4 v^2 \equiv 0 \pmod s,$$
$$ 3 u^2 \equiv - 4 v^2 \pmod s,$$
$$ \frac{3 u^2}{v^2} \equiv - 4 \pmod s,$$
$$ \frac{ u^2}{v^2} \equiv \frac{- 4}{3} \pmod s,$$
$$ \left(\frac{ u}{v}\right)^2 \equiv \frac{- 4}{3} \pmod s.$$
This says that the right hand side is a quadratic residue $\pmod s.$ It also means that $-12$ is a quadratic residue $\pmod s.$ This is false, though, as we have Legendre symbol
$$( -12 | s) = ( -3 | s) = ( -1 | s ) \cdot ( 3 | s).$$
Now, if $s \equiv 1 \pmod 4,$ we can ignore the $-1$ and switch the other pair. If $s \equiv 3 \pmod 4,$ then $( -1 | s ) = -1$ and $(3 | s) = - (s | 3)$.
In either case, we get
$$ (-3 | s) = (s | 3).$$
As $s \equiv 2 \pmod 3,$ we have
$$ ( s | 3) = ( 2 | 3) = -1.$$
So, in the end we have
$$( -12 | s) = -1,$$
and this contradicts the assumption that $v$ is nonzero $\pmod s.$ Thus, $v \equiv 0 \pmod s.$ From $ 3 u^2 + 4 v^2 \equiv 0 \pmod s,$ we then have $u \equiv 0 \pmod s.$ So $ s | \gcd(u,v)$ and $\gcd(u,v) \neq 1.$
This is a conjecture due to Dan Shanks or maybe Mullin. See the book of Paulo Ribenboim The new Book of Prime number records page 11.
Mullin, A. A., Bull. Amer. Math. Soc., 69, 737 (1963).
Shanks, D. Euclid's primes, Bull. Inst. Combin. Appl., 1, 33-36 (1991).
Look at page 5 of this link to Paulo Ribenboim's The Little Book of Bigger Primes.
Best Answer
The answer to the last question is yes. Much more is known: there is always a prime between $p$ and $2p$, by Bertrand's Postulate, which has long been a theorem.
The numbers $P_n=p_1p_2\cdots p_n +1\:$ have been looked at quite a bit, your $R_n$, for no clear reason, somewhat less. Very little of a general character is known about the $P_n$, and even less about the $R_n$. Prime numbers of either shape are called primorial primes.
It is not known whether there is an infinite number of primorial primes. More startlingly, it is not known whether an infinite number of the $P_n$ or $R_n$ are composite! There has been a great deal of computational work on these numbers, and primes do seem to become scarce among them as $n$ gets large. You can find some information here.