I think the answer should be yes. Here is a sketch. First, by the Bogomolov-Tian-Todorov theorem (as in the answer to the question about CY), the formal deformation space of a hyperkaehler $X$ is smooth. If we want to deform the polarization $L$, we have a nontrivial obstruction space $H^2(X, \mathcal{O}_X)\cong k$. Since $H^1(X, \mathcal{O}_X)=0$ (so that $Def(X,L)\to Def(X)$ is a closed immersion), geometrically, the formal deformation space $Def(X, L)$ is a hypersurface in the formal deformation space $Def(X)$ (see the argument in Deligne, P.
"Relèvement des surfaces K3 en caractéristique nulle"). To show that this hypersurface is smooth, we need to check that the "second order obstruction" product $H^1(X, T_X)\times H^1(X, T_X)\to H^2(X, \mathcal{O}_X)$ is non-degenerate (see the argument in Ogus "Supersingular K3 crystals"). Using the hyperkaehler structure $\omega$, this should be equivalent to $H^1(X, \Omega_X)\times H^1(X, T_X)\to H^2(X, \mathcal{O}_X)\cong k$ (cup product and trace) being non-degenerate.