Existing works on pre-computing ElGamal ephermal keys
It seems you want to decrypt the final value without revealing the private key. First, if someone knows the private key, they can issue a very simple non-interactive zero knowledge proof that the plaintext is a decryption of the ciphertext (the ciphertext being the accumulation of all the ballots) without revealing the actual value of the key. This is the standard way of approaching the problem. (I can add the proof if you are interested). 

Second, the approach of pre-committing to the random factors for Elgamal has been examined in the literature however for completely different reasons. There is a worry that if a voting machine chooses the randomness, they could try a few values until the ciphertext comes out with a certain pattern (e.g., the 5th bit is a 0 if the vote is for Alice and a 1 if it is for Bob). Your scheme incidentally avoids this. A more thorough approach is considered in On Subliminal Channels in Encrypt-on-Cast Voting Systems
.

If you insist on your approach, note that you never need to use $c_1$ at all. $c_2$ alone is a simpler primitive called a Pedersen commitment. The approach of encoding votes with Pedersen commitments, adding them up (under commitment), and then revealing the sum of the random factors is considered in Improving Helios with Everlasting Privacy Towards the Public. The difference is that voters choose the random factor and submit an encryption of it rather than using factors chosen for them. (It also is being done for totally different reasons: Pedersen commitments have a property called everlasting privacy).

In terms of the security of your approach, it is not clear who knows R and how voters get their value of R. For example, if I give voter 5 the value ($R_5$ - 1) and give voter 6 the value ($R_6$ +1), they will still add up to the correct value of $R$. However voter 5's message will be $m^x/k_1$ and voter 6's will be $m^x*k_1$. These may be sensible values for a vote: for example say that $k_1=m^2$. If voter 5 votes for 0, then this can be counted as a -1 and if voter 6 votes for 1, it can be counted as a +3. In other words, you end up with 2 votes for candidate 1 instead of one vote for each. 

It may be sufficient to choose $m$ in such a way that no one knows the discrete logarithm between $k_1$ and $m$ but I'd have to think about it some more before endorsing it.