The answer is 'no'. The generic pair $A$ and $B$ of $4$-by-$4$ Hermitian symmetric matrices will not have any nonzero real linear combination that has a double eigenvalue.
For a specific example, take $$ A = \begin{pmatrix}-3&0&0&0\\0&-1&0&0\\0&0&1&0\\0&0&0&3\end{pmatrix} \quad \text{and}\quad B = \begin{pmatrix}0&0&0&3i\\0&0&i&0\\0&-i&0&0\\-3i&0&0&0\end{pmatrix}. $$ Then $$ \det(aA+bB - tI_4) = (t^2-a^2-b^2)(t^2-3a^2-3b^2), $$ and the roots of this polynomial in $t$ are distinct unless $a=b=0$. (Recall that we are assuming that $a$ and $b$ are real, which, of course, implies that $t$ is real.)
Added Remark: To see the claim that this property holds for a generic linearly independent pair of Hermitian symmetric $4$-by-$4$ matrices $A$ and $B$, it is only necessary to observe the following: The question is whether, for a generic such pair $A$ and $B$ in the $16$-dimensional real vector space $\mathcal{H}_4$ consisting of $4$-by-$4$ Hermitian symmetric matrices, the (real) span of $A$, $B$, and $I_4$ contains a nonzero element of rank at most $2$. Now, it is not difficult to show that the cone $C_2$ of elements in $\mathcal{H}_4$ that have rank at most $2$ is a closed algebraic cone of dimension $12$ (one that is singular along the $7$-dimensional locus of elements of rank at most $1$). Hence the generic $3$-dimensional subspace of $\mathcal{H}_4$ will meet this cone only at the zero matrix. It is also an open (though not dense) condition on a $3$-dimensional subspace that it contain a positive definite element. Since the standard $\mathrm{GL}(4,\mathbb{C})$-action on $\mathcal{H}_4$ acts transitively on the space of positive definite elements, it follows that the generic pair $A$, $B$, together with $I_4$ will span a $3$-plane that meets $C_2$ only at the origin, which is what was to be shown.