Ratio of uncharged to charged side chains
I think you are simply misunderstanding the "uncharged to charged" part of the question. Imidazole is a base, so it gains a proton to become charged when the pH &lt; pKa. That's why you got the right answer when you used $\ce{HA+}$ as the acid, as you put it. In this case, unlike with most general chemistry problems of this nature, the acid is charged while the conjugate base is uncharged.

Below pH 6, imidazole exists largely as a protonated imidazolium ($\ce{HA+}$) species. Above pH 6, it is primarily in the neutral ($\ce{A}$) imidazole form. When imidazole is protonated (i.e., in acid or $\ce{HA+}$ form) it is charged. When it is deprotonated (i.e., in base or $\ce{A}$) form, it is uncharged.

So the proper expression is $$7 = 6 + \log\frac{[\ce{A}]}{[\ce{HA+}]}$$ which should lead you to the correct answer:

$$1 = \log\frac{[\ce{A}]}{[\ce{HA+}]}$$

$$10 = \frac{[\ce{A}]}{[\ce{HA+}]}$$

So, the ratio of uncharged:charged species is $\frac{[\ce{A}]}{[\ce{HA+}]}$, or 10:1.