The answer is yes, the theory of $L$ can be definable by a low-complexity definition quantifying over reals, even when $0^\sharp$ does not exist.
Here is one way to achieve this. Let me assume that the theory of $L$ is an element of $L$. This happens, for example, if $L_\kappa\prec L$ for some ordinal $\kappa$, because in this case the theory of $L$ is the same as the theory of $L_\kappa$, which is an element of $L$.
So let $t$ be the theory of $L$, which I have assumed is (coded by) a real in $L$. This real is therefore the $\alpha^{th}$ real in the $L$-order, and in order to define the theory $t$, it will suffice to define the ordinal $\alpha$.
Let $L[G]$ be a forcing extension of $L$ forcing to collapse $\aleph_{\alpha}^L$ to $\omega$. So in $L[G]$, the true $\omega_1$ is the same as $\omega_{\alpha+1}^L$, and we can determine this inside $H_{\omega_1}$. In that structure, we can define the class of ordinals that are cardinals in $L$, and there will be exactly $\alpha$ of them.
This makes the theory of $L$ definable inside $V=L[G]$ by a formula quantifying only over reals, and it will be $\Delta^1_n$ for some smallish $n$.
Let's try to find out how complex the definition is. My proposed definition is that in $V=L[G]$, the theory of $L$ is the theory coded by the real $t$ which is the $\alpha^{th}$ real in the $L$ order, where $\alpha$ is the number of infinite $L$-cardinals that are countable in $V$.
So, $t$ is as desired if there is a countable transitive model $L_\beta$ that thinks $t$ is the $\alpha^{th}$ real and which has exactly $\alpha$-many infinite cardinals, which do not get collapsed in any larger countable transitive $L_\gamma$, whereas all larger countable ordinals above those ordinals do get collapsed in some larger $L_\gamma$.
What is the complexity? It seems to be $\Sigma^1_4$. My initial thought that it might be $\Sigma^1_3$ are not right, as explained in the comments, since in that case it would be upward absolute by further forcing, but it clearly is not, since we could collapse more cardinals and thereby change the meaning of $\alpha$.