If you have a charge density different from 0 only within a finite region in space (mathematically speaking, ##\rho## has "compact support"), the assumptions of the theorem are fulfilled, because it's strictly 0 for ##r=\vec{x}>R## for some radius, ##R##. Then the solution of the Poisson equation (in SI units),
$$\Delta \Phi(\vec{x})=-\frac{1}{\epsilon_0} \rho(\vec{x})$$
is given by "summing up Coulomb potentials", i.e.,
$$\Phi(\vec{x})=\int_{B_R} \mathrm{d}^3 x' \frac{\rho(\vec{x}')}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|},$$
where ##B_R## is the solid sphere ("ball") of radius ##R## around the origin, which by definition contains all the charge there is.
Helmholtz's theorem in this form is for sure also valid under the conditions you quote, because if ##\rho(\vec{x}) = \mathcal{O}(r^{-\alpha})## for ##|\vec{x}|\rightarrow \infty## and ##\alpha>2##, then
$$\mathrm{d}^3 x' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|} = \mathrm{d} r' \mathrm{d\vartheta'} \mathrm{d} \varphi' \sin \vartheta' \mathcal{O}(r^{\prime -\alpha+1}),$$
i.e., the integrand falls faster with ##r'## than ##1/r##, and thus the integral over ##r'## converges for ##r' \rightarrow \infty##. The integrals over the angles is over finite regions (##\vartheta \in (0,\pi)##, ##\varphi \in (0,2 \pi)##) and thus don't diverge.
The theorem even works if ##\rho=\mathcal{O}(1/r^\beta)## with ##\beta>1##, because the potential is defined only up to an additive constant anyway. In this case you just subtract a clever constant, defining
$$\Phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(\vec{x}')}{4 \pi \epsilon_0} \left (\frac{1}{|\vec{x}-\vec{x}|}-\frac{1}{|\vec{x}_0-\vec{x}'|} \right).$$
The expression in the parentheses goes like ##\mathcal{O}(r^{\prime 2})## for ##r' \rightarrow \infty##, and thus the integral converges even under the more general condition on ##\rho##.
