This is not the case, and you can use Dyer-Lashof operations to see so. In the following I'll show this for $k = \Bbb F_2$ because that's the easiest case to compute. Bokstedt proved that $THH_*(\Bbb F_2) = \Bbb F_2[\sigma]$ where $|\sigma| = 2$. If it came from a commutative DGA, then the only nonzero Dyer-Lashof operations would be the squaring operations (because the operad action would factor through the projection down to the bottom class). We'll show that $Q^6 \sigma = \sigma^4$.

The description of $THH(k)$ by the cyclic bar construction makes it the derived smash product $k \wedge_{k \wedge^{\Bbb L} k^{op}} k$. The associated Kunneth spectral sequence comes from a simplicial commutative $k$-algebra, which gives the associated spectral sequence compatible Dyer-Lashof operations. It sits inside the Bokstedt spectral sequence (this is how Bokstedt computed the hidden extensions to show $THH_*(\Bbb F_2) = \Bbb F_2[\sigma]$). In particular, this spectral sequence takes the form of a Tor-spectral sequence
$$
Tor_{A_*} (\Bbb F_2, \Bbb F_2) \Rightarrow THH_*(\Bbb F_2)
$$
where $A_*$ is the dual Steenrod algebra $\Bbb F_2[\xi_i]$. The Tor-groups form an exterior algebra generated by $I/I^2$, where $I$ is the augmentation ideal of the dual Steenrod algebra. The exterior algebra is therefore generated by $[\xi_i]$.

However, Steinberger's calculations from the $H_\infty$ book tell us that
$$Q^6 \xi_1 =\xi_1^7 + \xi_1^4 \xi_2 + \xi_1 \xi_2^2 + \xi_3,$$
and therefore in $THH(k)$ we have an identity
$$Q^6 [\xi_1] = [\xi_3]$$
because everything else is decomposable. The spectral sequence degenerates and so $Q^6$ on the generator in degree two hits the only nonzero thing in degree 8.