Just a quick thought about showing that $3^n\alpha_n$ is integral. It's not too hard to prove via the Catalan number generating function that
\[\sum_{k=0}^{p-1} C_kx^k\equiv \frac{1-(1-4x)^{(p+1)/2}}{2x}-x^{p-1}\quad\bmod (p,x^p).\]
Now apply $D=x\frac{d}{dx}$ a bunch of times and set $x=1$ to get the sums you want. This will be in $\mathbb{Z}[1/6]$ but Then by using $(-3)^{(p-1)/2}\equiv \left(\frac{p}{3}\right)\bmod p$ you can control the 3-adic valuation (you get an extra 3 for each differentiation) and see that the only dependence on $p$ is the Legendre symbol $\left(\frac{p}{3}\right)$ which furthermore depends only on $p$ modulo 3. There's also a 2 in the denominator but I guess by looking a bit more carefully you can show this cancels out for $p$ not equal to 3. This is a bit sketchy but hopefully the idea is there.
In particular, letting Sage do the work, I get
\[\sum_{k=0}^{p-1}C_k\equiv \frac{{\left(\frac{p}{3}\right)}-1}{2}\quad\bmod p\]
\[\sum_{k=0}^{p-1}kC_k\equiv \frac{{-\left(\frac{p}{3}\right)}+1}{2}\quad\bmod p\]
\[\sum_{k=0}^{p-1}k^2C_k\equiv \frac{-{\left(\frac{p}{3}\right)}-3}{6}\quad\bmod p\]
which seem to fit your sequence as well as the results in the Zhi-Wei Sun paper.
I can't see how to get at the $\beta_j$ though.
Edited to add: it would appear that one can maybe do the same trick using
\[\sum_{k=0}^{p-1} C_kx^k\equiv \frac{1-(1-4x)^{(p^2+1)/2}}{2x}-px^{p-1}\quad\bmod (p^2,x^p)\]
to get the $\beta_j$ as well, though I'm not so convinced.
