Standard Deviation for Z-scores
In your example, you should calculate the probability that the value goes up, and use that as your $p$.  

That said, I think the calculation should be $\sigma = \sqrt{p(1-p)}$ otherwise, your standard deviation is inflated by a factor tied to $n$.  By using probabilities, you are essentially taking $n$ into account because:
$$p = {Number\ of\ "Up"\ instances \over n}$$
Where $n$ is the total number of observations.

Given the nature of the data, that is, it is binary, we would expect the $\sigma$ to be between 0 and 1 and not particularly sensitive to the number of observations.

That said, if we look at a few calculations, the value of $n$ in your formula significantly impacts the $\sigma$.

Assume for simplicity that $p$ is 0.5, that is an equal chance of going up or down.  If you have three different values of $n$ you get three vastly different values for $\sigma$.

Example 1: $n$ = 1 (Or not included)
$$\sigma = \sqrt{0.5 * (1-0.5)},\ \therefore \sigma = 0.5$$

Example 2: $n$ = 100
$$\sigma = \sqrt{0.5 * 100 * (1-0.5)},\ \therefore \sigma = 5$$

Example 3: $n$ = 1000
$$\sigma = \sqrt{0.5 * 1000 * (1-0.5)},\ \therefore \sigma = 15.8$$

Of the three, only example 1 really makes any sense intuitively. The mean of all instances would be 0.5 and you would expect the observed values to be 0.5 away from the mean.