Edit: If I understand correctly, Mike Shulman's answer claims that something like the following will be a semantic interpretation of propositional intuitionistic logic, as long as the meta-theory is intuitionistic.
Let $M$ be a function from the set of atomic propositions to the set $\{0,1\}$, and then interpret formulas recursively by: \begin{array}{lll} M \vDash p &= &M(p) = 1\\ M \vDash \top &= &\text{Triv.}\\ M \vDash \bot &= &\text{Abs.}\\ M \vDash \phi_1 \to \phi_2 &= & M \vDash \phi_1 \text{ implies } M \vDash \phi_2\\ M \vDash \phi_1 \lor \phi_2 &= & M \vDash \phi_1 \text{ or } M \vDash \phi_2\\ M \vDash \phi_1 \land \phi_2 &= & M \vDash \phi_1 \text{ and } M \vDash \phi_2 \end{array} This is my best attempt to adapt simple 2-value "classical" logic semantics to the situation of an intuitionistic meta-theory. The only changes I had to make was to avoid careless references to $\nvDash$, and to specify the meaning of implication separately, instead of defining it as $\neg \phi_1 \lor \phi_2$. I also changed the use of "iff" to equality to emphasize that I think we now need to think of the statements in a way that is closer to type theory. For example, for the meaning of an atomic proposition I think we are essentially saying "a proof of $M\vDash p$ is equivalent to a proof that $M(p) = 1$". For the meaning of $\bot$ we are saying "a proof of $M\vDash \bot$ is equivalent to a proof of absurdity". These two together mean that (defining $\neg \phi \equiv \phi \to \bot$) the meaning of $\neg p$ is essentially "a proof of $M \vDash \neg p$ is equivalent to a proof that $M \vDash p$ implies (meta-theoretic) absurdity". Now finally, since we are in intuitionistic meta-theory, we know that it is not given that for any atomic proposition $p$ we must have a proof of $M(p) = 1$ or a proof of $M(p) \neq 1$. This means that we can't conclude that there is always a proof of $M \vDash p \lor \neg p$ like we could in a classical meta-theory context.