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I've read in David Tong's lecture notes on gauge theory that the Hamiltonian of Yang-Mills theory does not depend on the angular parameter $theta$, because it can be absorbed in the electric field:

$$
mathcal{H}=frac{1}{g^2}text{tr}(mathbf{E}^2+mathbf{B}^2)=g^2text{tr}(mathbf{pi}-frac{theta}{8pi^2}mathbf{B})^2+frac{1}{g^2}text{tr}(mathbf{B}^2).
$$

Here, $g$ is the gauge coupling, $E_i=dot{A}_i$ is the non-Abelian electric field, $B_i=-frac{1}{2}epsilon_{ijk}F^{jk}$ the non-Abelian magnetic field, ${F}_{munu}$ is the gluon field strength, and
$$
mathbf{pi}=frac{partial mathcal{L}}{partial mathbf{dot{A}}}= frac{1}{g^2}mathbf{E}+frac{theta}{8pi^2}mathbf{B}
$$
is the momentun conjugate to $mathbf{A}$ (see pp. 39 and 40 of the lecture notes).

In contrast, it's well known that the Yang-Mills Lagrangian contains a topological $theta$-term:
$$
mathcal{L}= -frac{1}{2g^2}text{tr}(F^{munu}F_{munu})+frac{theta}{16pi^2}text{tr}(F^{munu}tilde{F}_{munu})=frac{1}{g^2}text{tr}(mathbf{dot{A}}^2-mathbf{B}^2)-frac{theta}{4pi^2}text{tr}(mathbf{dot{A}} mathbf{B}),
$$
where $tilde{F}_{munu}$ is the Hodge dual of $F_{munu}$, and the last equality holds for $A_0=0$ and $D_iE_i=0$.

How is this possible? Tong mentions that the $theta$-dependence in the Hamiltonian formalism is somehow hidden in the structure of the Poisson bracket, but he gives no detailed explanation. Why doesn't it appear in the Hamiltonian itself? The $theta$-term gives rise to the infamous strong CP problem, so how can we explicitly compute this $theta$-dependence of Yang-Mills in this formalism?

It's not particularly strange or unusual for a term to seemingly not appear in the Hamiltonian, but still have a physical effect. For example, the Hamiltonian for a free particle is
$$H = frac{p^2}{2m} = frac12 m v^2, quad p = mv.$$
On the other hand, the Hamiltonian for a particle in a magnetic field is
$$H = frac{(p-eA)^2}{2m} = frac12 m v^2, quad p = m v + e A.$$
They are of course different functions of $(x, p)$, reflecting the fact that the dynamics are different, but if you naively write the Hamiltonian in terms of the "physical" variables $(x, v)$ then you get the same result in both cases, since magnetic fields do no work. There is of course still an effect, because $v$ is related to $p$ in a different way. Your example with the $theta$-term is just a more complicated version of this, so there's nothing really weird to resolve.