As already indicated by Bjorn, this is hopeless. Whether or not a solution $\mu$ to a moment problem has an absolutely continuous part is decided exclusively at infinity; it does not depend on finitely many moments.

So you can always have two measures, one absolutely continuous and the other one purely singular, that have the same moments on an arbitrarily long initial piece. It is perhaps best to think of this in terms of the recurrence
$$
a_n p_{n+1}(z) + a_{n-1} p_{n-1}(z) + b_n p_n(z) = zp_n(z)
$$
satisfied by the polynomials orthogonal with respect to your measure. Knowing the first $N$ moments is the same as knowing the first $N$ coefficients $a_n,b_n$. (This also shows that it's easy to make the supports subsets of $[0,1]$, by controlling the $a$'s and $b$'s.)

So $\alpha_p=\infty$ for any $p>1$ and $\alpha_1=2s_0$. (Even if you insist that both measures are absolutely continuous, the same conclusions hold by approximation.)

**Some background information (digression):** (1) The collection of measures with (the first) $N$ prescribed moments can be described. Search for *Nevanlinna parametrization.*

(2) A related question that has a good answer is: Fix an interval $I\subset\mathbb R$ (or $I\subset [0,1]$ in your setting). What can we say about $\mu(I)$ if I'm given the first $N$ moments of $\mu$? There is classical work of Chebyshev and Markov on this. I wrote a paper on the continuous analog, which you are probably not interested in, but it also points to the relevant literature. See [item #16 here.][1]


  [1]: http://www2.math.ou.edu/~cremling/research/preprints.html