Xu's idea can be reproduced in arbitrary characteristic, without derivations, *with the extra hypothesis that the singularities are ordinary* (ie, $m_i$ distinct tangent directions at $p_i$), or at least each point has one direction of multiplicity one in the tangent cone. Without an assumption like this, I don't know if it can be done. **EDIT: It can't be done. See below.** Assume the base field $k$ is infinite; then the argument goes as follows.

Let $d$ and $m_1, \dots, m_n$ be such that for general $p_1, \dots, p_n$ there is an irreducible curve of degree $d$ with multiplicity $m_i$ at $p_i$. Fix such a set of points, and fix affine coordinates so that $p_1=(0,0)$ and $y=0$ is one of the tangent lines to the curve at $p_1$. Let $p(t)=(0,t)$ and for simplicity write $m=m_1$. By the assumption, for general $t\in k$ there is an irreducible curve of degree $d$ multiplicity $m$ at $p(t)$ and $m_i$ at $p_i$, $i=2,\dots, n$. Let $I\subset k[x,y]$ be the ideal of the points $p_2, \dots, p_n$ with their multiplicities, $I^e$ its extension to $k[x,y,t]$ and $J=I^e \cap (x,y-t)^{m}$. By hypothesis, there is an irreducible $F\in J$ of degree $d$ in $x$ and $y$.

Write $F=\sum_{i=0}^k t^iF_i$, where $F_i\in I$ are polynomials in $x,y$ of degree at most $d$, and $F_0$ has degree exactly $d$. Clearly $F\in J\subset(x,y,t)^m$ (a monomial ideal)
and it is easy to see that this implies $F_1 \in (x,y)^{m-1}$. If we show that $F_1\not\in (F_0)$, then the rest of Xu's argument works.

Since the multiplicity of $F_0$ at $p$ is exactly $m$, one can write $F=\sum_{j=0}^m G_j x^{m-j}(y-t)^{j}$, where $G_j$ are polynomials in $x,y,t$ and at least $G_1$ has nonvanishing constant term (by the choice of tangent line $y=0$). Then the summand for $j=1$ contributes a term $cx^{m-1}t$ to $F_1$ which can not be cancelled with any other term. This shows that the multiplicity at $p$ of $F_1$ is exactly $m-1$ and in particular that $F_1$ is neither zero nor equal to a scalar multiple of $F_0$.

**EDIT (added Oct 11):** Now consider the family of irreducible curves $C_{a,b}:(x-a)^2(y-b)^2=x^5+y^5$ which have a 4-fold point at the point $(a,b)$. For any parameterized curve $(a(t),b(t))$ in the plane $(a,b)$, expanding $F_t=(x-a(t))^2(y-b(t))^2-x^5-y^5=\sum_{i=0}^k t^iF_i$ gives $F_1=0$ in characteristic 2 and so the intersection-theoretic argument does not work (even after a linear change of variables, or for a germ of curve parameterized by two power series, the argument is the same).

Xu's argument is local in nature, and it has been used in other settings by many authors, always in characteristic zero (notably Lazarsfeld, see 5.2.3 in "Positivity in Algebraic Geometry I" and references there). I am quite sure that families of curves like the one above can be used to obtain counterexamples in positive characteristic on suitable surfaces. On the plane, however, Xu's conclusion is likely to be true, or at least counterexamples will be hard to construct (otherwise Nagata's conjecture would fail in positive characteristic, which is unknown and -as far as I am concerned- unexpected). In any case, Xu's argument can't be pushed to positive characteristic without some extra assumption on the singularities as above.