Isn't this just an application of Slutsky's Theorem? Write:

$$\frac{X - m}{s} = \frac{\sigma}{s}\frac{X - \mu}{\sigma} + \frac{\sigma}{s}\frac{\mu - m}{\sigma}$$

Then note that $\frac{\sigma}{s} \rightarrow 1$, $\mu - m \rightarrow 0$, and $\frac{X - \mu}{\sigma} \sim N(0,1)$.

EDIT: I guess you wanted a finite-sample result, not an asymptotic one. But if $n$ is large, the argument below shows why you'll be close to independent Normals.

Isn't this just an application of Slutsky's Theorem? Write:

$$\frac{X - m}{s} = \frac{\sigma}{s}\frac{X - \mu}{\sigma} + \frac{\sigma}{s}\frac{\mu - m}{\sigma}$$

Then note that $\frac{\sigma}{s} \rightarrow 1$, $\mu - m \rightarrow 0$, and $\frac{X - \mu}{\sigma} \sim N(0,1)$.