Let $D$ be a skew field with centre $K$ and maximal subfield $E$. Let $F$ be a finite extension of $K$ disjoint to $E$, that is, $F\cap E=K$. Is it true that the tensor product $D\otimes_K F$ has no zero divisors?
This is false in the complementary case: indeed, if $F\supseteq E$ then the tensor product will be a matrix ring since $E$ is a splitting field, and hence it will contain zero divisors.
First, note that although it is common in the literature to treat linear disjunction in a very casual way, in itself $F\cap E=K$ is not meaningful, as $F$ and $E$ are not both subfields of a bigger field. Or if they are, it should be explicit (lots of authors don't do it, and they are wrong because it can actually depend on the choices of embedding and not just on the isomorphism classes).
But in any case, no, what you are asking is not true: $D$ can have linearly disjoint splitting fields. For instance, if $D=(a,b)_K$ is a quaternion algebra, then you may take $E=K(\sqrt{a})$ and $F=K(\sqrt{b})$, which will be linearly disjoint if you take $a$ and $b$ to have different non-trivial square classes.