Beginning with the ordinary character table of the symmetric group $S_5$, one immediately gets the following Brauer characters in characteristic two:
$\begin{array}{c|c|c|c} S_5 & () & (12345) & (123) \\ \hline \beta_1 & 1 & 1 & 1\\ \beta_2 & 4 & -1 & 1 \\ \beta_3 & 5 & 0 & -1 \\ \end{array}$
It is easy to see that both $\beta_1$ and $\beta_2$ are irreducible. Moreover, the third (and last) irreducible 2-Brauer character must be given either by $\beta_3$ or by $\beta_3 - \beta_1$. Indeed, by a computer search, I found a representation $S_5 \to \mathrm{GL}(4,2)$ affording the Brauer character $\beta_3 - \beta_1$. So $\beta_3$ is not irreducible.
Is it possible to come to the same conclusion by character theoretic arguments?
I actually found a purely character theoretic solution to my question, which requires some work though. I am still very interested in seeing other solutions!
Starting with the character table of the alternating group $A_5$ it is possible to derive the table of its irreducible 2-Brauer characters in a straightforward way (this is carried out here).
$\begin{array}{c|c|c|c|c} A_5 & () & (12345) & (13524) & (123) \\ \hline \gamma_1 & 1 & 1 & 1 & 1\\ \gamma_2 & 2 & \varphi-1 & \psi -1 & -1 \\ \gamma_3 & 2 & \psi-1 & \varphi-1 & -1\\ \gamma_4 & 4 & -1 & -1 & 1\\ \end{array}$
($\varphi$ is the golden ratio, and $\psi$ is its algebraic conjugate)
At this point, it is easy to see that $\beta_3-\beta_1 = (\gamma_2)^{S_5}$ is induced by a Brauer character of $A_5$, and so is a Brauer character as well.