Have you tried plotting this using ggplot2 in R? It has a nice semi transparency feature with the Cairo package which makes guesstimating the mean for such residual plots easy. For example you could have each point semi transparent and you could visually check if they are centered around 0. But overall by looking at the image you posted, no reason to think otherwise.
You can do this on a plot of residuals - plain, raw residuals, not standardized or studentized - vs anything at all - residuals vs fitted, residuals vs x (indeed any predictor in a multiple regression), residuals vs index number, residuals vs a variable you didn't use in the regression, whatever you like. It's only the fact that the residuals ($y-\hat{y}$) are plotted on one of the axes that's important.
The residuals are most typically plotted on the y-axis so in that case it's the y-axis you pay attention to. It doesn't make a difference what's on the x-axis when doing this, you don't pay any attention to it. [If your residuals have been plotted on the x-axis for some reason then that is the direction you pay attention to.]
Here's a plot of residuals vs fitted values for regression on a particular data set. There are 50 observations (though one of the residuals near 0 is not visible because it lays exactly over another point). Since we want an interval containing 2/3 of the points, we'll have 1/6 of the points above the top of the interval and 1/6 of the points below it.
Since 50/6 is 8-and-a-bit, let's leave 8 points outside either end, so we'll mark our lines about halfway between the 8th and 9th points from the top and bottom of the plot. [You could mount an argument for drawing the lines right at the 9th point and saying that's 8.5 points outside each of the lines and 50-8.5-8.5 = 33 between them, but this is pretty approximate anyway so I won't worry too much. For example, 2/3 is slightly too small so I'll stick with 8 points outside each end of the interval.]
Measuring the distance between those two horizontal lines gives a distance of about 2.2 in this case. Since that interval containing the middle 2/3 of the data should be about 2 standard deviations wide, we estimate the standard deviation to be $s\approx 2.2/2 = 1.1$.
The actual standard deviation of residuals from the regression output ($\sqrt{SSE/(n-2)}\,$) was $s = 1.102$.
Not too shabby!
Best Answer
A dependent mixture model (hidden Markov model) may be of use, depending on the type of deviations expected.
Assume that your observations come from two distributions (or states), both of which are normally distributed, but have different mean and variance.
A number of parameters can be estimated: The initial state probabilities (2 parameters), the state transition probabilities between neighbouring data points (4 parameters) and finally the mean and variance of the two distributions (4 parameters).
In R, this model can be estimated using the depmixS4 package:
See http://cran.r-project.org/web/packages/depmixS4/vignettes/depmixS4.pdf for references