Is multiplicative blinding less secure than additive?
In a finite field $\mathbb{F}_q$, both maskings are perfectly secure, provided that $x \neq 0$ for the multiplicative masking.

This is easy to see.  The finite field $\mathbb{F}_q$ defines two groups: the additive group $\mathbb{F}_p^+$ (i.e., $\mathbb{F}_q$ equipped with addition) and the multiplicative group $\mathbb{F}_q^*$ (i.e., $\mathbb{F}_q \setminus \{0\}$ equipped with multiplication).

Proposition  Let $\mathbb{G}$ be a group and let $\star$ denote the group law in $\mathbb{G}$.  Given $x \in \mathbb{G}$, if $r$ is a uniformly random element in $\mathbb{G}$ then so is $y:=x \star r$.

Let $x \in \mathbb{F}_q$.  We have:


If $r$ is drawn uniformly at random in $\mathbb{F}_q^+$ then $x+r$ (where $+$ denotes the addition in $\mathbb{F}_q$) is uniformly random in $\mathbb{F}_q^+$;
If $r$ is drawn uniformly at random in $\mathbb{F}_q^*$ then $x*r$ (where $*$ denotes the multiplication in $\mathbb{F}_q$) is uniformly random in $\mathbb{F}_q^*$, provided that $x \neq 0$ (that is, $x \in \mathbb{F}_q^*$).