This is not an answer to my question. This is more like a comment, I want to see if I understand BCnrd's argument or if I miss the point completely. I still feel shaky with scheme theory. Feel free to make any comments.

So suppose that $X$ is as in the original question and assume furthermore (only for the sake of simplicity) that it is affine. Then we have that

$X=Spec(A)$ where $A=\mathbf{C}[x_1,\ldots,x_n]/(f_1,\ldots,f_r))$ where the $f_i$'s are polynomials in the $x_i$'s. Note that $A$ is a noetherian ring. Now let $\mathcal{F}$ be the $O_{\overline{X}}$-coherent sheaf of module on $\overline{X}$ which is obtained from GAGA. As BCnrd pointed out $\mathcal{F}$ is also simultaneously an $O_{\overline{X}}$-sheaf
of algebra. In particular, it is a **finite type**
$O_{\overline{X}}$-sheaf of algebra. From now on we will only use the fact that it is 
of finite type as a sheaf of algebra (if you prefer quasi-coherent over an noetherian space). Since $\mathcal{F}$ is a finite type sheaf of $O_{\overline{X}}$-algebra and $X$ is affine then $\mathcal{F}|X$ is generated by sections over $X$ (this is an easy exercise but important to point out). So in other words one has that 

$B:=\mathcal{F}(X)=A[y_1,y_2,\ldots,y_m]$ for $y_1,\ldots,y_m\in \mathcal{F}(X)$. Now let $Y_1,\ldots, Y_m$ be formal variables and let $R:=A[Y_1,\ldots,Y_m]$. Note that $R$ is also an noetherian ring.

We have a natural map $\phi:R\rightarrow B$ which takes $Y_i\mapsto y_i$. Since $R$ is noetherian it follows that $ker(\phi)$ is finitely generated so let 
$ker(\phi)=(g_1,\ldots,g_t)$ where the $g_i$'s are polynomials in the $Y_i$'s and coefficients in $A$.

Now let $B:=A[Y_1,\ldots,Y_m]/(g_1,\ldots,g_t)$ so that
$$
B=\mathbf{C}[x_1,\ldots,x_n,Y_1,\ldots,Y_m]/(f_1,\ldots,f_r,g_1,\ldots,g_t)
$$
where we think now of the $g_i$'s as being polynomials in the $x_1,\ldots,x_n,Y_1,\ldots,Y_m$.

Now one has to verify that the "analytification" of the $\mathbf{C}$-scheme structure of $Spec(B)$ gives a space which is isomorphic (as an analytic variety) to $Y$ and compatible with the map $f$ (this is probably not that difficult to prove). Now because of the smoothness of $X$ and the fact that $f:Y\rightarrow X^{an}$ was analytically unramified it is (probably) easy to see that $Spec(B)$ is smooth. (the smoothness is equivalent to the regularity of the local ring which (I guess) may be detected after completion of the local ring).

So it seems that the only place where the coherence was used was at the outset on the
analytic sheaf $\mathcal{F}^{an}$ of $O_{\overline{X}}$-module on the $\mathbf{C}$-projective space $\overline{X}$. This allowed us to apply **GAGA** in order to get the existence of an **algebraic sheaf of $O_{\overline{X}}$-algebra $\mathcal{F}$ on $\overline{X}$**. And from there we only used the fact that $\mathcal{F}|X$ was a finite type sheaf of $O_{X}$-algebra.

Please let me know if I miss something important.