I get why $\sqrt{9} = \pm 3$. But (at least I think) the ± is there because there's a certain ambiguity as to which number was squared to obtain $9$.
Does that mean that if we remove the ambiguity $\sqrt{3^2} = 3$ ?
One argument could be that since $\sqrt{3^2} = \sqrt{9} = \pm 3$. Then again we could argue that we know for a fact that $9$ is the result of squaring the number $3$ and should therefore be $\sqrt{3^2} = 3$.
I apologize as I'm only a beginner and this may perhaps seem too basic.
When we have $\sqrt{x}$ it is usually assumed to be the principal square root operator, which means it returns only the positive root. We have $$\sqrt{\cdot}: \mathbb R_{\geq 0} \to \mathbb R_{\geq 0}\\ x \mapsto \sqrt{x}$$
In your case, by definition
$$\sqrt{x^2} =\left|x\right| = \begin{cases} x & \mbox{ if } x > 0 \\ -x & \mbox{ if } x < 0 \\ 0 & \mbox{ if } x = 0\end{cases}$$
Your line of thinking makes sense, but it's not exactly like that - it's not that we "don't know" which value was squared to get it; rather, both are answers.
In most (almost all) contexts, $\sqrt n$ refers to only the positive value of the square root. So, $\sqrt{3^2}$ would just be $3$, but so would $\sqrt{(-3)^2}$. In this situation, you have to be careful because $\left(\sqrt n\right)^2=n$ but $\sqrt{n^2}=|n|$.
If you think about it, $\sqrt n$ is just "roots of the equation $x^2=n$" (hence why it's called square root) and so there will be two answers (well, except when $n=0$, then there's only one distinct answer).
Overall, your line of thinking makes sense, but it's not that we "don't know the original value". Depending on your definition, either only the positive value is correct, or occasionally both values are right, and either way, you cannot count on $\sqrt{n^2}$ to equal $n$.
