First of all, there is a priori no reason to assume that $a(t)$ and $a^\dagger(t)$ satisfy the canonical commutation relation for $t>0$. But that is probably not a very satisfying answer to your question, so let me motivate why operator generally cannot maintain their commutation relations at times $t>0$.

The Lindblad equation describes relaxation to equilibrium, and therefore the system state in the Schrödinger picture generically* converges towards a unique steady state $\rho_\infty$ at long times, $\rho(t) \to \rho_\infty$, independent of the initial state $\rho(0) = \rho_0$. A direct consequence is that, if $A$ is some operator, the expectation value $\operatorname{tr}[A\, \rho(t)]$ converges to a number $A_\infty$ independent of $\rho_0$.

Translating that to the Heisenberg picture, we find $$ \lim_{t \to \infty} \operatorname{tr}[A(t)\, \rho_0] = \lim_{t \to \infty} \operatorname{tr}[A\, \rho(t)] = A_\infty , $$ which is only possible if $$ \lim_{t \to \infty} A(t) = A_\infty\, \mathbb I . $$ Here, $\mathbb I$ is the identity operator. Operators in the Heisenberg picture must therefore converge to a multiple of the identity, and their commutator will go to zero.

* Not every Lindblad equation has a unique steady state, but almost all Lindblad equations that we as Physicists are interested in do. There are mathematical criteria for that. In any case, it is not important for the argument above that all Lindblad equations are like that.

First of all, there is a priori no reason to assume that $a(t)$ and $a^\dagger(t)$ satisfy the canonical commutation relation for $t>0$. But that is probably not a very satisfying answer to your question, so let me motivate why operators generally cannot maintain their commutation relations at times $t>0$.

The Lindblad equation describes relaxation to equilibrium, and therefore the system state in the Schrödinger picture generically* converges towards a unique steady state $\rho_\infty$ at long times, $\rho(t) \to \rho_\infty$, independent of the initial state $\rho(0) = \rho_0$. A direct consequence is that, if $A$ is some operator, the expectation value $\operatorname{tr}[A\, \rho(t)]$ converges to a number $A_\infty$ independent of $\rho_0$.

Translating that to the Heisenberg picture, we find $$ \lim_{t \to \infty} \operatorname{tr}[A(t)\, \rho_0] = \lim_{t \to \infty} \operatorname{tr}[A\, \rho(t)] = A_\infty , $$ which is only possible if $$ \lim_{t \to \infty} A(t) = A_\infty\, \mathbb I . $$ Here, $\mathbb I$ is the identity operator. Operators in the Heisenberg picture must therefore converge to a multiple of the identity, and their commutator will go to zero.

* Not every Lindblad equation has a unique steady state, but almost all Lindblad equations that we as Physicists are interested in do. There are mathematical criteria for that. In any case, it is not important for the argument above whether all Lindblad equations are like that.