In complex analytic geometry the theorem of Fischer and Grauert states: A smooth family of compact complex manifolds is locally trivial if and only if all fibres are analytically isomorphic.
Hence in complex analytic geometry there are no examples - with smooth fibre - of the type you are searching for.
Could you please detail your statement about étale cover of curves?
The statement you have written is not correct, for several reasons.
There are two related (true) theorems, which I'll state now. (Both can be found in Shafarevich Chapter II --- I don't have the book to hand, so I can't give precise references.)
Theorem A: Let $f: X \dashrightarrow \mathbf A^1$ be a rational function on a smooth variety $X$. If $X$ is regular along every codimension-1 subvariety $Y \subset X$, then $X$ is regular.
Sketch of proof: Everything is local on $X$, so we can assume $X$ is affine, embedded in some $\mathbf A^n$. Then $f=g/h$ for some polynomials $g$ and $h$ with no common factor. The indeterminacy locus of $f$ is then the zero locus of $h$, which is either empty or codimension 1 in $X$. $\ \square$
Theorem B: Let $f: X \dashrightarrow \mathbf P^m$ be a rational map from a smooth variety $X$ to projective space. Then the indeterminacy locus of $f$ has codimension at least 2 in $X$.
Sketch of proof: Let's stick to the case $m=1$ for simplicity of notation; the general case works in the same way. Reducing to the affine case as above, we can write $f$ as $[g,h]$ where $g$ and $h$ are polynomials. If $g$ and $h$ have a common factor we can remove it, so again we may assume they don't. The indeterminacy locus of $f$ is $\{g=0\} \cap \{h=0\}$, which therefore has codimension at least 2 in $X$. $\ \square$
It seems as though you somehow mixed up these two theorems.
Let me close with a couple of remarks:
Theorems A and B almost seem to be in conflict with each other! The key point to notice is that in Theorem B we are talking about a map to projective space. Roughly speaking, in this case there are "more places" for the codimension-1 subsets of $X$ to map to, hence less indeterminacy.
Neither Theorem A nor B is true without some hypotheses on the singularities of $X$. (I assumed "smooth" for simplicity; you can check that "normal" would also suffice.) For example, let $C$ be a nodal cubic curve and $n: \mathbf P^1 \rightarrow C$ its normalisation. Then Theorem B fails for the rational map $n^{-1}$. (A counterexample for Theorem A is a bit harder to find!)
Update: This addresses the extra questions that the OP added later. If the linear system $|L|$ decomposes into mobile and fixed part as $|M|+F$, then we extend it to all codimension-1 points by simply "forgetting" about $F$. That is: outside $F$, the maps given by $|L|$ and $|M|$ are the same, so the map given by $|M|$ extends that given by $|L|$. Since $|M|$ is mobile, it has no fixed components, so the corresponding map only fails to be defined in codimension 2.
(I note that none of this is specific to surfaces. One thing that is specific to surfaces is that a mobile linear system has finite base locus, and such a complete linear system is actually semi-ample (some multiple is basepoint-free) by the Zariski--Fujita theorem.)
In your example of the pencil of conics on a cubic surface, the pencil $|M|$ of conics is basepoint-free, so the corresponding map is actually a morphism extending your original rational map. (To see that the pencil of conics is actually basepoint-free, let $C$ be a conic in the pencil. Then $C+l$ is a plane section of the cubic, so $(C+l)^2=3$. On the other hand $C \cdot l = 2$ by Bezout, and $l^2=-1$, so we get $C^2=0$. Hence two distinct conics in the pencil are disjoint.)
Best Answer
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:
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! :)