My guess would be that what you are stating is true. You can even be sure that for $\alpha =2$ the limit will be finite.
Here's my reasoning. Pick your point O, and your ray. Then for $r$ sufficiently large, the sphere centered at $O$ and of radius $r$ contains all the point charges (since there is a finite number of them, they must lie in a finite region of space). If you want to be more explicit denote by $R$ the distance of the furthest point charge from $O$, then you can pick $r>R$.
Then, when $r\rightarrow \infty$, the details of the point particle distribution inside the sphere $R$ disappear. In other words, when looking from afar, your sphere $R$ just looks like a point particle with total charge the sum of the charges of the point particles. Hence, the electric field produced by your distribution, when $r\rightarrow \infty$, should be $E\propto \frac{\sum q}{r^2}$. Of course, if the charges cancel, we could have some faster decreasing field, but it is sure that it will decrease at least as fast as $r^{-2}$.
Let me be more specific. Consider the total electric field produced at our faraway point p :
$$\vec{E(p)} \propto \sum_i \frac{q_i}{|\vec{r}-\vec{r}_i|^2}\vec{e}_{i}$$
Where $\vec{r}$ denotes the position of $p$, $\vec{r}_i$ the positions of the different charges measured from O, and $\vec{e}_i = \frac{\vec{r}-\vec{r}_i}{|r-r_i|}$.
Then, using the triangle inequality :
$$|E(p)|\leq \sum_i \frac{q_i}{|\vec{r}-\vec{r}_i|^2}$$
We now expand using the fact that $\vec{r}_i/r \rightarrow 0$, i.e. $r$ is really big :
$$|E(p)|\leq \sum_i \frac{q_i}{r^2}(1+2\frac{\vec{r}\cdot\vec{r}_i}{r^2}+O(\left(\frac{r_i}{r}\right)^2))$$
So, finally, $$lim_{r\rightarrow \infty}|E(p)|r^2 \leq \lim_{r\rightarrow \infty}\sum_i\frac{q_i}{r^2}r^{2} = \sum_i{q_i}<\infty$$
Concerning your comment, as you can see for two opposite charges, the limit is 0, meaning the norm of the magnetic fields decreases faster than $r^{-2}$ ($r^{-3}$, as you remarked).
Let's try and see if it is possible to determine the exponent of $r$ that we need in order to get a finite non-zero constant. First of all, we should exclude the case of no charges, since that clearly gives 0.
Let's start expanding the electric field without using the triangle inequality, since we want an exact limit. I will use $\vec{e}_r = \frac{\vec{r}}{r}$
$$\vec{E}(p) = \sum_i \frac{q_i}{|r-r_i|^3}(\vec{r}-\vec{r}_i) = \sum_i\frac{q_i}{r^2}(\vec{e}_r-\frac{\vec{r}_i}{r})(1+3\frac{\vec{e}_r\cdot\vec{r_i}}{r}+O(\frac{r_i^2}{r^2}))$$
Now, notice in the expansion above that no matter to which order we push the expansion, it is always rational exponents that we add. This is simply due to the fact that $\frac{1}{1-q} = \sum_i^\infty q^i$. The expansion at arbitrary order will thus be a function containing only monomials of the form $r^{-\alpha}$, where $\alpha$ is an integer. (I believe that's the multipole expansion people are referring to, it's been a long time for me so I'm not too sure).
Now, when $r\rightarrow \infty$, we only care about the lowest order term of the expansion. To illustrate, we first look at the first term, $\frac{|\sum_i q_i|}{r^2}$. If it is non-zero, then $\alpha = 2$ and we are done. If $|\sum_i q_i| = 0$ (example, the dipole) , then we go to next order, $r^{-3}$ and so on.
I believe this does prove that you can't have an non-integer exponent in $r$ for the expansion, let alone an irrational one.
Now there is however one last problem : what happens if the expansion is 0 for any order ? That would mean that the electric field is exactly 0 at infinity. Intuitively I would say this should be impossible to do with a finite number of charges, since we are essentially imposing an infinite number of constraint, but I did not find anyway to prove it rigorously yet.
In any case, it should not invalidate the claim that the exponents are integers.
