Is Normal(mean, variance) mod x still a normal distribution? Is the following distribution a still normal one? 

The range of values is constrained to hard limits $  \{0, 255\}  $. And it's generated by $  \mathcal{N}(\mu, \sigma^2)  \operatorname{mod} 256 + 128^{\dagger} $. So it's not that I'm simply truncating a normal, but folding it back into itself. Consequently the probability is never asymptotic to 0. 
Normal, or is there a more appropriate term for it? 

$\dagger$ The +128 is just to centre the peak otherwise it looks silly as $\mu = 0$. $\sigma \sim 12; < 20$. I expect that $\mu$ can be ignored for the purposes of classification.
In detail, it's a plot of 128 + (sample1 - sample2) & 0xff. I wouldn't fixate on the centring too much as there's some weird +/- stuff to do with computer networking and transmission of unsigned bytes.
 A: *

*No, by definition your distribution is not normal. The normal distribution has unbounded support, yours doesn't. QED.

*Often enough, we don't need a normal distribution, only one that is "normal enough". Your distribution may well be normal enough for whatever you want to use it for.

*However, I don't think a normal distribution is really the best way to parameterize your data. Specifically, even playing around with the standard deviation, there doesn't seem to be a way to get the kurtosis your data exhibits:

Note how we either get all the mass over a much smaller part of the horizontal axis for small SDs, or get much more mass in the center of the distribution than in your picture for large SDs.
mu <- 128
ss <- c(5,10,20,40)
nn <- 1e7

par(mfrow=c(2,2),mai=c(.5,.5,.5,.1))
for ( ii in 1:4 ) {
    set.seed(1)
    hist(rnorm(nn,mu,ss[ii])%%256,col="grey",
        freq=FALSE,xlim=c(0,256),xlab="",ylab="",main=paste("SD =",ss[ii]))
}

I also tried a $t$ distribution by varying the degrees of freedom, but that didn't look very good, either.
So, if your distribution is not normal enough for your purposes, you may want to look at something like a Pearson type VII distribution. If you truncate this using the modulo operator, it will again not be a Pearson VII strictly speaking, but it may again be "sufficiently Pearson VII".
