Yes, it is. There are standard criteria (stated for example in Ken Ono's book "Web of Modularity") that indicate when an eta quotient is modular on $\Gamma_{0}(N)$, and these show that $\eta(12\tau)^{2}$ is a modular form of weight $1$ on $\Gamma_{0}(144)$ with Nebentypus $\chi_{-4}$ (with order of vanishing $1$ at every cusp, somewhat surprisingly).
For $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, define $g(z) | M = \det(M)^{1/2} (cz+d)^{-1} f\left(\frac{az+b}{cz+d}\right)$ to be the weight $1$ "slash" operator.
If we let $f(z) = \eta(\tau)^{2}$, then $f(z) | \begin{bmatrix} 12 & 0 \\ 0 & 1 \end{bmatrix}$ transforms nicely under $\Gamma_{0}(144)$, and hence $f(z)$ transforms nicely under $\begin{bmatrix} 12 & 0 \\ 0 & 1 \end{bmatrix} \Gamma_{0}(144) \begin{bmatrix} 1/12 & 0 \\ 0 & 1 \end{bmatrix}$. This contains $\Gamma(12)$.