Can Reed Solomon parity blocks be used as an all-or-nothing transform? Consider the following scheme:


Perform an (N,N) Reed-Solomon encoding (i.e. N data blocks, N parity blocks)
Drop the N data blocks and keep only the N parity blocks.


Are these N parity blocks an all-or-nothing transform? (meaning missing even one parity block does not permit decoding the original data blocks)
Reed-Solomon codes are Maximum-Distance-Separable codes and thus have the property that in a $[n,k]$ RS code ---- meaning than there are a total of $n$ blocks of which $k$ are data blocks and $n-k$ are parity block ($n=2N$ and $k=N$ in the OP's notation) ---- any $k$ blocks are sufficient to reconstruct the entire codeword, that is, all the blocks can be found if we know $k$ of them. Note that it does not matter whether the $k$ blocks available are a mixture of data and parity blocks or parity blocks only, just as long as $k$ blocks are available. 

So, in the scheme envisioned by the OP, if one has a $[2N,N]$ Reed-Solomon code ($N$ data blocks, $N$ parity blocks, for a total of $2N$ blocks) and keeps only the $N$ parity blocks, the $N$ data blocks can be reconstructed. If fewer than $N$ of the parity blocks, then the data blocks cannot be reconstructed in the sense that the decoder can come up with a list of $M$ possible candidates for the $N$ data blocks but cannot say which of the candidates is the correct one.  Here, each of the $M$ candidates on the list is a vector $[D_1, D_2, \ldots, D_N]$
of $N$ data block (each $D_i$ is a data block) and the decoder cannot determine which is the correct vector of $N$ data blocks. What is $M$?  Well, $M$ is the number of possible distinct values that each data block might have and thus is either the number of elements in the finite field over which the Reed-Solomon code is defined, or some power $L$ thereof if each data block is a vector of length $L$ over the finite field (and we are using an interleaved Reed-Solomon code). Either way, $M$ is pretty large.