Let $\phi$ be a self-financing strategy that replicates a time $T$ option payoff $X$ on stock $S$. By definition of a trading strategy, $\phi$ is previsible. Finally, let $V_t$ be the time $t$ value of the portfolio implementing $\phi$.
Usual theorem: If $S_te^{-rt}$ is a $\mathbb{Q}$-martingale then $V_t = \mathbb{E}_\mathbb{Q}[e^{-r(T-t)}X|\mathcal{F}_t]$.
Note if the price of the option were anything other than $V_t$, we would have arbitrage.
My Claim: If $S_t$ is a $\mathbb{Q}$-martingale then $V_t = \mathbb{E}_\mathbb{Q}[X|\mathcal{F}_t]$.
No discounting necessary, and again this $\phi$ is a replicating self-financing strategy, so the value of the option must again be $V_t$ for all $t$.
Proof of My Claim. Consider discrete time and let $S_t$ be a $\mathbb{Q}$-martingale. By definition of a self-financing strategy, $\Delta V_{t+1} = \phi_{t+1}\Delta S_{t+1}$. where $\Delta V_{t+1} = V_{t+1} - V_t$. Hence \begin{align*} \mathbb{E}_\mathbb{Q}[\Delta V_{t+1}|\mathcal{F}_t] =\mathbb{E}_\mathbb{Q}[\phi_{t+1}\Delta S_{t+1}|\mathcal{F}_t] =\phi_{t+1}\mathbb{E}_\mathbb{Q}[\Delta S_{t+1}|\mathcal{F}_t] = 0, \end{align*} where the second equality is because $\phi$ is previsible. So $V_t$ is a $\mathbb{Q}$-martingale, and \begin{align*} V_t = \mathbb{E}_\mathbb{Q}[V_T | \mathcal{F}_t] = \mathbb{E}_\mathbb{Q}[X | \mathcal{F}_t], \end{align*} where the second equality is because $\phi$ replicates $X$.
What am I missing? Why do we only consider discounted stock prices, and hence what's the point of discounting the expectation? The only good reason I can think of for considering only discounted stock prices is that martingale representation theorem guarantees the existence of self-financing strategies in this case. But still, the usual theorem is valid for any self-financing strategy, so the discounting still seems unnecessary.
Update: Okay, my previous example indeed assumed I was trading in only one stock, in which it's generally not possible to have a replicating self-financing strategy (right?). Let me prove a new, more general claim, again asserting discounting is not necessary by trading in two stocks $S^1$ and $S^2$ and requiring them both to be $\mathbb{Q}$-martingales. This will be slightly different than the "usual" theorem above in that we'll require both stocks to be martingales, but this seems valid. Furthermore, if we had an option payoff that was a function of two stocks, we could use this same claim to trade in only those two stocks; i.e., no "extra" stocks needed to trade in.
Again let $\phi$ be a self-financing strategy that replicates a time $T$ option payoff $X$ on stock $S^1$, but now $\phi$ trades in both $S^1$ and $S^2$. By definition of a trading strategy, $\phi$ is previsible. Finally, let $V_t$ be the time $t$ value of the portfolio implementing $\phi$.
Updated claim: If the vector $S = (S_1, S_2)$ is a $\mathbb{Q}$-martingale; i.e., each component of $S$ is a $\mathbb{Q}$-martingale, then $V_t = \mathbb{E}_\mathbb{Q}[X|\mathcal{F}_t]$.
Proof. Let $S = (S_1, S_2)$ be a $\mathbb{Q}$-martingale. By definition of a self-financing strategy, $\Delta V_{t+1} = \phi_{t+1}\cdot\Delta S_{t+1}$, where $\Delta V_{t+1} = V_{t+1} - V_t$ and "$\cdot$" is the dot product. Hence \begin{align*} \mathbb{E}_\mathbb{Q}[\Delta V_{t+1}|\mathcal{F}_t] =\mathbb{E}_\mathbb{Q}[\phi_{t+1}\cdot\Delta S_{t+1}|\mathcal{F}_t] =\phi_{t+1}\cdot\mathbb{E}_\mathbb{Q}[\Delta S_{t+1}|\mathcal{F}_t] = 0, \end{align*} where the second equality is because $\phi$ is previsible. So $V_t$ is a $\mathbb{Q}$-martingale, and \begin{align*} V_t = \mathbb{E}_\mathbb{Q}[V_T | \mathcal{F}_t] = \mathbb{E}_\mathbb{Q}[X | \mathcal{F}_t], \end{align*}
Does this seem correct? If so, it indeed seems discounting (in general, by some numeraire) would be necessary for the "usual" theorem, but this seems just as valid.