Let me try to answer the modified question. (I assume we are working over an algebraically closed field of characteristic zero.)
The answer to your question depends crucially on what is meant by ruled. (I think this was a source of confusion in the comments.) There are two different defintions I know of, neither one standard:
- $S$ is geometrically ruled (what Hartshorne calls ruled) if $p: S \rightarrow C$ has every fibre isomorphic to $\mathbf{P}^1$;
- $S$ is birationally ruled (what some other people called ruled) if $S$ is birational to a product $C \times \mathbf{P}^1$.
Now Noether--Enriques tells you that a rational fibration is birationally ruled (at least once you know there are some smooth fibres, which you get from Bertini's second theorem) but you might worry that there are some singular fibres that stop it from being geometrically ruled. However, this can't actually happen, for the following reason. By your definition of rational fibration, every fibre of $p$ is an irreducible reduced rational curve, but any such curve which is not isomorphic to $\mathbf{P}^1$ must be singular, hence have arithmetic genus greater than 0; however, this cannot happen for a flat family such as $p: S \rightarrow C$. (See Hartshorne for justifications of all these assertions.)
So a rational fibration is the same thing as a geometrically ruled surface. Just to be sure that these are really different from the birationally ruled surfaces, let's give an example: take $p: S \rightarrow C$ a geometrically ruled surface, and let $\pi: S' \rightarrow S$ be the blowup of any point on $S$. Then $p \circ \pi: S' \rightarrow C$ is a birationally ruled surface, but one of its fibres is a reducible curve, so it is not geometrically ruled.
Actually, this last example also highlights a problem with your idea in the final paragraph: blowing up points unfortunately can't turn a fibration with singular fibres into one with smooth fibres, because it introduces extra components into some of the fibres.
I hope this helps.
Edit: Finally I think I can answer the intended question. Let $S$ be a birationally ruled surface, not isomorphic to $\mathbf{P}^2$. Contract $(-1)$-curves on $S$ until we reach a minimal surface $S_m$. So we have a morphism $p: S \rightarrow S_m$. Now according to Enriques' classification, there are three possibilities for $S_m$:
It is a geometrically ruled surface $\pi: S_m \rightarrow C$ over a curve of genus $>0$;
It is a Hirzebruch surace $\pi: S_m \rightarrow \mathbf{P}^1$;
It is $\mathbf{P}^2$.
The first two types are geometrically ruled, so $\pi \circ p$ gives a morphism to a curve with generic fibre $\mathbf{P}^1$. (We get $S$ from $S_m$ by blowing up a finite number of points, so that doesn't affect the generic fibre.)
The only problem is if we arrive at $\mathbf{P}^2$. But in that case we just stop contracting curves at the penultimate step, to get a morphism $p: S \rightarrow \Sigma_1$ where $\Sigma_1$ is the blowup of $\mathbf{P}^2$ in one point. Now $\Sigma_1$ is also a geometrically ruled surface, so we can argue as in the other cases.
And if that doesn't answer the question, I quit! :)
Lemma 9.15 of Gathmann shows that the blow-up of $X$ at some functions $f_1, ..., f_r\in A(X)$ depends only on the ideal $(f_1, ..., f_r)\unlhd A(X)$.
This shows that if we have an isomorphism $f: X\to Y$, and if the ideal $J\unlhd A(Y)$ is sent to $I\unlhd A(X)$ under the induced isomorphism $f^*: A(Y)\to A(X)$, then the blow-up of $X$ at $I$ will be isomorphic to the blow-up of $Y$ at $J$.
We can show with the Affine Jacobi Criterion (Prop. 10.11) that the only singular points are $0$ in both $X_k$ and $X_l$. In our case this is equivalent to the tangent space having dimension at least 2 (since the local dimension is always $1$ for these irreducible curves). Since the dimension of the tangent space depends solely on the stalk $\mathscr{O}_{X,a}\cong \mathscr{O}_{Y,f(a)}$, we know it must be preserved under isomorphisms. We conclude if $f:X_k\to X_l$ is an isomorphism then the origin goes to the origin.
This means $I = I(\{0\})\unlhd A(X_l)$ is sent to $f^*I = I(\{f(0)\})\unlhd A(X_k)$.
Therefore, the blow-up $\tilde{X_k}$ of $X_k$ at the origin, i.e., at $I(\{0\})$, is isomorphic to the blow-up $\tilde{X_l}$ of $X_l$ at the origin, i.e., at $I(\{f(0)\})$.
Best Answer
The structure theory is more complicated than for curves, but is still relatively concrete:
Birational maps of surfaces factor as sequences of blow-ups at points.
Every birational equivalence class of surfaces, other than the ruled surfaces (those birational to $C \times \mathbb{P}^1$ for $C$ a curve) contains a unique minimal model, which is to say a smooth surface $S$ that cannot be "blown down": any rational map $S \to S'$ is an isomorphism. (Ruled surfaces also have minimal models, but they are not unique.)