You should say what bundle you are taking Chern/Segre classes of. (As diverietti points out, knowledge of the two is equivalent.) Some people use "Chern class of $X$" to mean "Chern class of the tangent bundle to $X$." If this is what you mean, then no, because you haven't incorporated the data of the embedding.

Alternatively, you could mean the classes of $\mathcal{O}(1)$ restricted from projective space. In this case, there is only one nonzero Chern class, $c_1$, and the Segre classes are $s_k = (-c_1)^k$. So all you can see is $\int_X (c_1)^{\dim X}$, better known as $\deg X$. This is equivalent to knowing the first term of the Hilbert polynomial, but no more.

If you know $c_1(\mathcal{O}(1))$ and you know the Chern (or Segre) classes of $T_X$, then yes! This is the Hirzebruch-Riemann-Roch formula.

**ADDED** We have a short exact sequence of vector bundles on $X$: $0 \to T_X \to T_{\mathbb{P}^n}|_X \to N_{X/\mathbb{P}^n} \to 0$. So $c(T_X) c(N_{X/\mathbb{P}^n}) = c(T_{\mathbb{P}^n})$ (where $c$ is the total Chern class). Remembering that $s(E) c(E) = 1$, with $s$ the total Segre class, we have
$$c(T_X) = c( T_{\mathbb{P}^n}) s(N_{X/\mathbb{P}^n}).$$
Writing $H$ for the hyperplane class $c_1(\mathcal{O}(1))$, we have $c(T_{\mathbb{P}^n}) = (1+H)^{n+1} - H^{n+1}$. So knowing the degree of $X$ and the Segre class of the normal bundle determines the Chern classes of the tangent bundle. There may be some further simplification that can be performed, but I don't see it right now.