No. Unitarity guarantees that whenever the Hamiltonian exists, energy is a real number, but it does not imply energy conservation.
Writing
$$|\psi(t)\rangle = U(t,t_0)|\psi(t_0)\rangle\text{,}$$
we should have the composition law
$$U(t+t',t_0) = U(t+t',t')U(t',t_0)\text{.}$$
If $U$ is translation-invariant, $U(t+t',t_0+t') = U(t,t_0)$, then it follows that $U(t+t',0)$ $=$ $U(t+t',t')U(t',0)$ $=$ $U(t,0)U(t',0)$, so we can take $U(t,0) = A^t$ and use the fact that the logarithm of a unitary operator is skew-hermitian whenever it exists: $A = e^{-iH/\hbar}$ for some constant, hermitian $H$.
If it's not time-translation-invariant, we can't do this. Just as with classical mechanics, energy conservation and time translation are linked. However, for short time-translations $\delta t$,
$$U(t+\delta t,t_0) = U(t,t_0) + (\delta t)\dot{U}(t,t_0) + {\mathcal O}(\delta t^2)\text{,}$$
and unitarity requires that, to first order in $\delta t$,
$$\begin{eqnarray*}
1 &=& \left[U(t,t_0) + (\delta t)\dot{U}(t,t_0)\right]\left[U(t,t_0) + (\delta t)\dot{U}(t,t_0)\right]^\dagger \\
&=&1 + (\delta t)\underbrace{\left[\dot{U}U^\dagger + U\dot{U}^\dagger\right]}_0\text{.}
\end{eqnarray*}$$
Thus the operator $\dot{U}U^\dagger$ is skew-hermitian, so defining $H = i\hbar \dot{U}U^\dagger$, $H$ must be hermitian, and in particular
$$\begin{eqnarray*}
U(t+\delta t,t_0) &=& U(t,t_0)U^\dagger(t,t_0)U(t,t_0)\\
&=& \left[1+(\delta t)\frac{H}{i\hbar}\right]U(t,t_0)\text{.}
\end{eqnarray*}$$
Re-arranging and taking the limit as $\delta t\to 0$,
$$i\hbar\frac{\partial}{\partial t}U(t,t_0) = HU(t,t_0)\text{,}$$
which is the Schrödinger equation for the time-evolution operator. To usual Schrödinger equation follows immediately by multiplication by the state ket.
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On the flip-side, energy conservation, in the sense of a time-independent Hamiltonian, doesn't guarantee unitarity. For a trivial example, take a nonrelativistic particle with Hamiltonian $\hat{\mathcal H} = \frac{1}{2m}{\hat{p}}^2 + V(x)$ with complex but time-independent potential. Then probability is not conserved, as (for the right imaginary sign) the particle has an exponentially decaying probability of being around.