There is a general necessary condition, and then there are *ad hoc* sufficient conditions. The necessary condition is related to the Brauer-Manin obstruction. Via pullback of Brauer classes, there is a pairing,
$$X(K)\times \text{Br}(X) \to \text{Br}(K), \ (x,\alpha) \mapsto x^*\alpha.$$
Said differently, there is a set map,
$$ X(K)\to \text{Hom}_{\mathbb{Z}}(\text{Br}(X),\text{Br}(K)).$$
If $x,y\in X(K)$ are $R$-equivalent, then they are in the same fiber of this set map. Thus, if there exists $\alpha\in \text{Br}(X)$ such that $x^*\alpha$ does not equal $y^*\alpha$, then $x$ and $y$ are not $R$-equivalent.

On the other hand, proving that points are $R$-equivalent is much trickier. Regarding cubic surfaces, one important reference is the following.

MR1875181 (2003c:11070) Reviewed

Swinnerton-Dyer, Peter

Weak approximation and R-equivalence on cubic surfaces. (English summary)

Rational points on algebraic varieties, 357–404,
Progr. Math., 199, Birkhäuser, Basel, 2001.

11G35 (11G30 14G05)

For trying to prove direct $R$-equivalence, the simplest guess is that there is a singular plane cubic in the cubic surface that contains $x$ and $y$. If $L$ is the line spanned by $x$ and $y$, then either $L$ is contained in $X$ (in which case you are done), or else projection of $X$ away from $L$ defines an elliptic fibration over a complementary $\mathbb{P}^1$. In this case, there is a degree $12$ divisor in $\mathbb{P}^1$ that is the discriminant of the fibration. If that degree $12$ divisor happens to have a rational point, then you are usually done: the normalization of the corresponding singular plane cubic will be what you want. The one caveat is that the singular set of that curve may contain $x$ or $y$, in which case you may not yet be done: the inverse image of that point in the normalization may be a closed point whose residue field has degree $2$ over $K$.