Yes. Though the energy will not be unbounded, but bounded from above, if my calculation is correct.
For real scalar field under $(+---)$ metric, besides the negative classical kinetic energy for the Lagrangian $$\mathcal{L}=-\frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi - \frac{1}{2} m^2 \phi^2 \tag{1} $$,
the classical equation of motion will be
$$ (\square - m^2 )\phi=0 . \tag{2}$$
For plane wave $\phi ~\sim e^{ipx} $, it gives $p^2+m^2 = (p^0)^2 - \mathbf{p}^2+m^2=0$ which is inconsistent with relativistic energy momentum relation. I am not sure if it is necessary to quantize it.
Though the energy-momentum-relation argument will not work for the Dirac field, we can quantize it to see the energy will be negative definite.
$$\mathcal{L} = \bar{\psi}( -i \gamma^{\mu} \partial_{\mu} - m ) \psi \tag{3}$$
The classical equation of motion is
$$ (i \gamma^{\mu} \partial_{\mu} +m) \psi=0 \tag{4}$$
To preserve all properties of $u(\mathbf{p})$ and $v(\mathbf{p})$, we define
$$ \psi =: u(\mathbf{p}) e^{ipx} $$
$$ \psi =: v(\mathbf{p}) e^{-ipx} $$
Thus we can replace the $u(\mathbf{p})$ as $v(\mathbf{p})$ and $v(\mathbf{p})$ as $u(\mathbf{p})$ in the expansion of $\psi$ and $\bar{\psi}$.
By
$$\pi = -i \bar{\psi} \gamma^0 $$
then
$$H= \int d^3 x \bar{\psi} ( i \gamma^i \partial_i \psi + m ) \psi $$
Plug in expansions of spinors in the Schrodinger picture
$$
\psi = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( a_{\mathbf{p}}^s v^s (\mathbf{p}) e^{-i\mathbf{p} \cdot \mathbf{x} } + b_{\mathbf{p}}^{s\dagger} u^s(\mathbf{p}) e^{i \mathbf{p} \cdot \mathbf{x} } \right)
$$
$$
\bar{\psi} = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( b_{\mathbf{p}}^s \bar{u}^s (\mathbf{p}) e^{-i\mathbf{p} \cdot \mathbf{x}} + a_{\mathbf{p}}^{s\dagger} \bar{v}^s(\mathbf{p}) e^{i\mathbf{p} \cdot \mathbf{x}} \right)
$$
we have
$$ H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} a_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(\mathbf{p}) $$
$$ = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} a_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(\mathbf{p}) $$
$$ = \sum_s \int \frac{ d^3p}{ (2\pi)^3} E_{\mathbf{p}} ( b_{\mathbf{p}}^{s} b_{\mathbf{p}}^{s\dagger} - a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) $$
$$ = \sum_s \int \frac{ d^3p}{ (2\pi)^3} - E_{\mathbf{p}} (b_{\mathbf{p}}^{s\dagger} b_{\mathbf{p}}^{s} + a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) - \infty $$
Changing anticommutator into commutator will make the spectrum unbounded.
