No, this is generally not true.
If $E_1=E_2=E$, we have $\langle\Psi|H|\Psi\rangle=\langle\Psi|E\Psi\rangle=E\langle\Psi|\Psi\rangle=E$, which is not $E_1+E_2$ unless $E=0$.
If $E_1\ne E_2$, then if $H$ is Hermitian, $\psi_1$ and $\psi_2$ are orthogonal:
$\langle\psi_1|H|\psi_2\rangle=\langle\psi_1|H\psi_2\rangle=\langle\psi_1|E_2\psi_2\rangle=E_2\langle\psi_1|\psi_2\rangle$
and
$\langle\psi_1|H|\psi_2\rangle=\langle H\psi_1|\psi_2\rangle=\langle E_1\psi_1|\psi_2\rangle=E_1\langle\psi_1|\psi_2\rangle\;,$
and thus $\langle\psi_1|\psi_2\rangle=0$ if $E_1\ne E_2$.
Thus the mixed terms in the expectation value vanish, and we have
$ \begin{eqnarray} \langle\Psi|H|\Psi\rangle &=& \langle\psi_1|H|\psi_1\rangle+\langle\psi_2|H|\psi_2\rangle+\langle\psi_1|H|\psi_2\rangle+\langle\psi_2|H|\psi_1\rangle \\ &=&\langle\psi_1|H|\psi_1\rangle+\langle\psi_2|H|\psi_2\rangle \\ &=& \langle\psi_1|\psi_1\rangle E_1+\langle\psi_2|\psi_2\rangle E_2 \\ &=& \frac{|\psi_1|^2E_1+|\psi_2|^2E_2}{|\psi_1|^2+|\psi_2|^2} \;, \end{eqnarray} $
where I've put $|\psi_1|^2+|\psi_2|^2=1$ into the denominator to show more explicitly that this is a weighted average of $E_1$ and $E_2$. This result holds quite generally: If $\Psi$ is a normalized sum of components which are eigenvectors of a Hermitian operator $H$, then the expectation value $\langle\Psi|H|\Psi\rangle$ is the weighted average of the eigenvalues, weighted by the squared amplitudes of the components. In the present case, this may happen to be equal to $E_1+E_2$ if these two don't have the same sign, but it usually wouldn't be.