This is more an extended comment than an actual answer. If I'm not mistaken, then deciding whether a given graph has a strict unfriendly partition is NP hard, i.e. there probably is no easy-to-decide characterisation. Here is a reduction from 3-SAT to this decision problem.

Given a 3-SAT formula $F$, define a graph $G$ as follows:

- Take a path on 5 vertices $(a,u,b,v,c)$.
- For every variable $x_i$ of $F$ take a path on 4 vertices $(t_i,u_i,v_i,f_i)$, and attach to every $t_i$ and every $f_i$ as many leaves as there are clauses in $F$.
- for every clause $C_j$ of $F$ take a vertex $y_j$ and connect this vertex to $t_i$ if $x_i$ appears in $C_j$, and to $f_i$ if $\lnot x_i$ appears in $C_j$. Furthermore, add edges from $a$, $b$, and $c$ to every $y_j$.

We claim that $G$ has a strict unfriendly partition if and only if $F$ is satisfiable.

Assume that $G$ has a strict unfriendly partition. The vertices $u$,$v$,$u_i$, and $v_i$ make sure that $a,b,c$ end up in the same part whereas $t_i$ and $f_i$ end up in different parts. Denote the part containing $a,b,c$ by $A$.

Since 3 of the 6 neighbours of any $y_j$ are contained in $A$, we know that $y_j \notin A$. Furthermore, at least one of the remaining neighbours must be in $A$, otherwise we wouldn't have a strict unfriendly partition. By construction of $G$, this shows that every clause is satisfied if we set $x_i = \mathrm{true}$ for $t_i \in A$ and $x_i = \mathrm{false}$ for $f_i \in A$, and thus $F$ is satisfiable.

Conversely, assume that $F$ is satisfiable and let $x_i = \hat x_i$ be an assignment of values that satisfies $F$. Define a partition of $V(G)$ by

$$A = \{a,b,c\} \cup \{t_i, v_i, \text{leaves attached to }f_i \mid \hat x_i = \mathrm true\} \cup \{f_i, u_i, \text{leaves attached to }t_i \mid \hat x_i = \mathrm false\},$$
and $B = V(G) \setminus A$. This is strictly unfriendly in $a,b,c,u,v,u_i,v_i$ because all their neighbours are in in the other part, respectively. It is strictly unfriendly in $t_i$ and $f_i$ because the leaves attached to them outweigh the neighbours of the form $y_j$. It is strictly unfriendly in $y_j$ because $a,b,c \in A$ and $x_i = \hat x_i$ was assumed to satisfy $F$ (and thus at least one more neighbour of $y_j$ is contained in $A$).