This is true. Actually, even better, these singularities will be terminal and Gorenstein, so as mild as it can get. Well, at least if we assume that you are working over an algebraically closed field, but otherwise you would have to be more careful about what exactly do you mean by these assumptions and questions, so I'll assume that that's what you meant.

Your assumptions imply that all the fibers are Gorenstein. Since $\mathbb P^2$ is smooth, this implies that then $X$ is Gorenstein.
Furthermore, your assumptions imply that all the fibers are $1$-dimensional, so $f$ is equidimensional, and hence it is flat by "miracle flatness". So, as Ariyan noted, then $f$ is smooth at every $x\in X$ such that $x\neq x_p$ and hence $X$ is smooth at all of those points.
This also implies that $X$ is normal (it is $S_2$ since it is Gorenstein and its singular locus has at most codimension $2$, so it is also $R_1$).

Now, looking at one of these $x_p$'s, we still have that $X$ is Gorenstein and we know that a complete intersection curve through this point is a simple node. As an exercise, try to prove directly (say via a direct local computation) that this implies that then $X$ is canonical (of index $1$) at these points.

If you get stuck, then use this argument:
Let $C_1$ and $C_2$ be two smooth curves in $\mathbb P^2$ that intersect transversally at $p$ and let $D_i=f^*C_i\subseteq X$ for $i=1,2$. (For the record, the $D_i$ are reduced divisors on $X$.) Now, $D_1\cap D_2 =f^{-1}(p)$, which is a nodal curve, so it has slc singularities.
Then by inversion of adjunction (applied twice) $(X,D_1+D_2)$ is log canonical. Now if you scrape away the $D_i$, this actually means that $X$ is terminal at the points of the form $x_p$. (Note that $(D_1, D_1\cap D_2)$ is also log canonical, so $D_1$ is canonical. It is locally isomorphic to a cone over a quadric.)