Assume the continuum hypothesis fails, so that we have some intermediate cardinal $d$ with $\omega < d < c$. Choose a random subset $S \subset [0,1]$ of cardinality $d$ by sampling $d$ i.i.d. uniform points.
Intuitively, it seems like $S$ will be nonmeasurable with probability one. That is, almost all intermediate cardinality subsets of $[0,1]$ are nonmeasurable. This is because its outer measure is 1, and the outer measure of the complement is 1, both with probability 1.
Unfortunately, I am unsure whether this claim can be formalized. The random process that generates $S$ is just a product of random variables, so that part is easy. However, it's not obvious that the set of measurable sets is itself measurable, without which the claim can't be stated.
Can this claim be appropriately formalized into a true theorem?
First of all, it is consistent with $ZFC+\neg CH$ that every subset of $[0,1]$ of cardinality less than $\mathfrak{c}$ is measurable. This follows from Martin's axiom, for instance.
However, even if not all such sets are measurable, the statement you are asking about will never be true. As you guessed, the issue is being able to measure the event in question at all. You are considering (the completion of) the product $\sigma$-algebra on the sample space $[0,1]^d$, and every element of the product $\sigma$-algebra depends on only countably many factors. That is, if $A\subseteq[0,1]^d$ is in the product $\sigma$-algebra, there is a countable subset $C\subset d$ such that $A=A_0\times[0,1]^{d\setminus C}$ for some $A_0\subseteq[0,1]^C$.
Now, if the event of $S$ being nonmeasurable were to have probability $1$, it would need to contain some set $A$ in the product $\sigma$-algebra which also has probability $1$. But in fact, this event contains no nonempty sets in the product $\sigma$-algebra at all. Indeed, if it contained such a nonempty set, that would mean that there would be some countable subset $S_0$ of $[0,1]$ such that every subset $S$ of cardinality $d$ containing $S_0$ is nonmeasurable. This is impossible (since you can always add $d$ elements of the Cantor set to $S_0$ to get a measurable set, for instance).
Note that assuming not all sets of cardinality $d$ are measurable, the same argument applies to the event that $S$ is measurable. So in such a model, the event of $S$ being measurable is indeed nonmeasurable: neither it nor its complement contains any nonempty set in the product $\sigma$-algebra!