Given $u$ is a classic solution for $$-\varepsilon \, \Delta u + u_t + H(x,Du) = 0$$ and $\varphi\in C^2$ (called a super solution?) satisfies $$-\varepsilon \, \Delta \varphi + \phi_t + H(x,D\phi) \geq 0$$ with the same initial condition $u(x,0) = \phi(x,0) = f(x)$.
Can I say $u(x,t)\leq \varphi(x,t)$ with the only assumption that $H(x,p)$ is bounded uniformly continuous on $\mathbb{R}^n\times [0,T]$ and coercive in $p$?
The answer is no, without some additional growth constraints on $u$ and $\phi$. You can just take $H\equiv 0$, and then you have the heat equation, which has infinitely many smooth solutions for the same initial data if you do not impose growth constraints.
If $u$ and $\phi$ are bounded, then the answer will be yes. The trick is to modify either the sub or supersolution in such a way that $u \leq \phi$ outside of a large rectangle $[-N,N]\times [0,T]$. Then use the usual maximum principle.
Generally, when you prove uniqueness of viscosity solutions to HJ equations ($\varepsilon=0$), you assume the solutions are bounded.
EDIT: Let me add a few words about the trick I mentioned above. For $\lambda >0$ write
$v(x) = \phi(x) + \lambda^2 \log(1+|x|^2) + \lambda t.$
If $H$ is uniformly continuous in $\nabla u$, then for $\lambda$ sufficiently small
$-\varepsilon \Delta v + v_t + H(\nabla v, x) > 0.$
Furthermore, if $\phi$ and $u$ are bounded, then $u \leq v$ on $\partial B(0,R)\times [0,T]$ for sufficiently large $R$. Use the ordinary maximum principle for bounded domains to get $u \leq v$ on $\mathbb{R}^n\times [0,T]$. Then send $\lambda \to 0^+$. The argument only requires uniform continuity of $H$ in $\nabla u$.