In fact, we always have equality.

Suppose $v \in \text{Eig}(H_\mu, \mu)$.  Write $v = v_1 + v_2$ where 
$v_1 = H_1 v$, $v_2 = (I-H_1) v$ are orthogonal.
We have
$$ \mu v_2 = H_\mu v - \mu v_1 = \mu H_1 v + (1-\mu) H_2 v - \mu v_1 =  (1-\mu) H_2 v$$
i.e. $$H_2 v = \frac{\mu}{1-\mu} v_2$$
Now since $H_2$ is idempotent, $H_2 v_2 = v_2$, so that 
$$H_2 v_1 = H_2 v - v_2 = \dfrac{2\mu - 1}{1-\mu} v_2 $$
But by symmetry,
$$0 = v_1^T H_2 v_2 = v_2^T H_2 v_1 = \dfrac{2\mu - 1}{1-\mu} v_2^T v_2$$
and since $\mu \ne 1/2$, $v_2 = 0$, and $v = v_1 \in \text{img}(H_1) \cap \text{ker}(H_2)$.  

EDIT: For larger convex combinations, an analogous statement is not true.  For example, consider $H_\mu = \dfrac{1}{6} H_1 + \mu_2 H_2 + \left(\dfrac{5}{6} - \mu_2\right) H_3$ for the  symmetric idempotents
$$ \eqalign{H_1 &= \pmatrix{1 & 0\cr 0 & 0\cr},\cr H_2 &= \pmatrix{9/34 & -15/34\cr -15/34 & 25/34\cr}, \cr H_3 &= \pmatrix{25/26 & -5/26\cr
-5/26 & 1/26\cr}} $$
The condition for $H_\mu$ to have eigenvalue $1/6$ is $8712 \mu_2^2 - 9108 \mu_2 + 2125 = 0$, which has a root near $\mu_2 = .3514749310$.
Then you can check that for this $\mu_2$, $H_\mu$ has an eigenvector for eigenvalue $ 1/6$
which is not in $\text{img}(H_1)$ or $\text{ker}(H_2)$ or $\text{ker}(H_3)$.