How to determine and derive the protonation state of a functional group based on its pKa?
You're on the right track.  The Henderson-Hasselbalch equation is where this rule of thumb comes from; the key thing to remember is that, in biochemistry, your $\mathrm{pH}$ is fixed by the environment (there's always something acting as a buffer).  This is in contrast with typical H.H. problems in general chemistry where the amount of molecule and its $K_\mathrm{a}$ is changing the $\mathrm{pH}$ of the solution.

Anyhow, given that $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ are fixed values, we're free to simply compare the ratio of $[\ce{A-}]$ and $[\ce{HA}]$:

\begin{align}
\mathrm{pH} &amp;= \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}\\
\log\frac{[\ce{A-}]}{[\ce{HA}]} &amp;= \mathrm{pH} - \mathrm{p}K_\mathrm{a}\\
\frac{[\ce{A-}]}{[\ce{HA}]} &amp;= 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}\\
\end{align}

If $\mathrm{p}K_\mathrm{a} = \mathrm{pH} + 1$ then we have $\frac{[\ce{A-}]}{[\ce{HA}]} = 10^{-1}$.  In other words, we have 10 times as much $\ce{HA}$ as we have $\ce{A-}$ (i.e., we have $\approx91\%\ \ce{HA}$) as we should expect for something that is mildly basic.

If $\mathrm{p}K_\mathrm{a} = \mathrm{pH} - 1$ then we have $\frac{[\ce{A-}]}{[\ce{HA}]} = 10^{1}$.  In other words, we have 10 times as much $\ce{A-}$ as we have $\ce{HA}$ (i.e., we have $\approx91\%\ \ce{A-}$) as we should expect for something that is mildly acidic.

Clearly, if the $\mathrm{pH}$ differs from the $\mathrm{p}K_\mathrm{a}$ by only 1 $\mathrm{pH}$ unit, we don't really have a completely protonated or deprotonated molecules, but it's certainly mostly protonated or deprotonated.  As the $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ get further apart, then we continue to change this ratio by orders of magnitude.

Let's look at a concrete example.  Physiological $\mathrm{pH}$ is 7.4 and the side chain of aspartic acid has a carboxylic acid group with a $\mathrm{p}K_\mathrm{a} = 3.9$, so we predict

$$\frac{[\ce{RCOO^-}]}{[\ce{RCOOH}]} = 10^{7.4 - 3.9} = 10^{3.5} = 3200$$

In other words, we still have some protonated sidechain, but we have >3000 times more deprotonated sidechain than protonated sidechain (did you notice that? the difference was >3 so we have >1000 difference in concentration ratio).  For all practical purposes, the aspartic acid side chain is "completely deprotonated" at physiological $\mathrm{pH}$.