The function $x(\tau) = j(\tau)^{1/3}$ is a hauptmodul, just not for the group that you indicate. This function is also invariant under $\tau \mapsto \frac{2 \tau + 1}{\tau + 1}$ and $\tau \mapsto \frac{\tau+1}{\tau + 2}$, and this means that $x(\tau)$ is a hauptmodul for an index $3$ subgroup of ${\rm SL}_{2}(\mathbb{Z})$, which is the normalizer of a non-split Cartan modulo 3. (This has been known for quite a while. If you want to read a proof of this, see the paper of Imin Chen titled "On Siegel's modular curve of level 5 and the Class Number One Problem" published in the Journal of Number Theory in 1999. Look in Section 4.) The subgroup you specify has index at least $18$ (and it may not even be congruence).

I think that the natural modification of your second question also has a negative answer. In particular, $$ \frac{x(\tau)^{2}}{x(\tau) - 12} = q^{-1/3}(1 + 12q + 144q^{2} + 1976q^{3} + \cdots) $$ is a modular function for the same group as $x(\tau)$ and also has non-negative Fourier coefficients. That's because this function can be written as $\frac{1}{j(\tau) - 1728} \cdot (x(\tau)^{4} + 12 x(\tau)^{3} + 144 x(\tau)^{2})$. The latter factor has non-negative Fourier coefficients and the former does too by thinking carefully about the identity $\frac{1}{j(\tau) - 1728} = \frac{E_{4}^{3}/E_{6}^{2} - 1}{1728}$. Here $E_{4} = 1 + 240 \sum_{n=1}^{\infty} \sigma_{3}(n) q^{n}$ and $E_{6} = 1 - 504 \sum_{n=1}^{\infty} \sigma_{5}(n) q^{n}$ are the usual Eisenstein series.

The function $x(\tau) = j(\tau)^{1/3}$ is a hauptmodul, just not for the group that you indicate. This function is also invariant under $\tau \mapsto \frac{2 \tau + 1}{\tau + 1}$ and $\tau \mapsto \frac{\tau+1}{\tau + 2}$, and this means that $x(\tau)$ is a hauptmodul for an index $3$ subgroup of ${\rm SL}_{2}(\mathbb{Z})$, which is the normalizer of a non-split Cartan modulo 3. (This has been known for quite a while. If you want to read a proof of this, see the paper of Imin Chen titled "On Siegel's modular curve of level 5 and the Class Number One Problem" published in the Journal of Number Theory in 1999. Look in Section 4.) The subgroup you specify has index at least $18$ (and it may not even be congruence).

I think that the natural modification of your second question also has a negative answer. (Edited to fix the example.) In particular, $$ x(\tau) \frac{j(\tau)}{j(\tau)-1728} = q^{-1/3} (1 + 1976q + 2133020q^{2} + \cdots) $$ is a modular function for the same group as $x(\tau)$ and also has non-negative Fourier coefficients. With some thought, one can see that $j(\tau)$ has non-negative Fourier coefficients, and so does $\frac{1}{j(\tau) - 1728} = \frac{E_{4}^{3}/E_{6}^{2} - 1}{1728}$. Here $E_{4} = 1 + 240 \sum_{n=1}^{\infty} \sigma_{3}(n) q^{n}$ and $E_{6} = 1 - 504 \sum_{n=1}^{\infty} \sigma_{5}(n) q^{n}$ are the usual Eisenstein series.