Is multiplicative blinding less secure than additive? It's easy to see that additive blinding (e.g., $x+r$ for secret x and random r) is perfectly secure in a finite field (this is a one-time-pad) and statistically secure for $r$ uniformly distributed in a domain sufficiently larger than $x$.

What I'm confused about is multiplicative blinding (i.e., $x \cdot r$) - in some places, I read that it's being used and is as secure as additive blinding, while in others it's stated that it is much less secure (e.g., multiplicative blinding less secure). Assuming, of course, both $x \neq 0$ and $r \neq 0$, is multiplicative blinding as good as additive (ignoring side channel attacks), and if not - what's the explanation?
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, $x \in \mathbb{F}_q^*$).