I don't know how to reduce a standard oracle to a phase oracle, but the proof can actually be generalized to the standard oracle model, with a little more work. You can check out the proof on my blog for better math typesetting.
Notations and definitions
Let $S=\{0,1\}^n$ be the search space, and $R\subset S$ be the solutions, we define $N=|S|,M=|R|$. Let $f_R:S\to\{0,1\}$ be a function that checks whether a given solution is valid, i.e. $f_R(x)=1\iff x\in R$. To find a solution, we are allowed to query $f_R$ as an oracle in our algorithm design. For a quantum algorithm, we are given an "oracle operation" $O_R$ such that $O_R|x\rangle|b\rangle=|x\rangle|b\oplus f_R(x)\rangle$
First, we introduce a lemma to bound the sum of vector norms:
Lemma 1
Let $\{\alpha_i\}$ and $\{\beta_i\}$ be some finite sequence of vectors in a inner product space. Then we have
$$\begin{align}
\sum_i\|\alpha_i+\beta_i\|^2&\le\left(\sqrt{\sum_i\|\alpha_i\|^2}+\sqrt{\sum_i\|\beta_i\|^2}\right)^2\\
\sum_i\|\alpha_i+\beta_i\|^2&\ge\left(\sqrt{\sum_i\|\alpha_i\|^2}-\sqrt{\sum_i\|\beta_i\|^2}\right)^2
\end{align}$$
Proof of Lemma 1
By the triangle inequality, we have
$$\begin{align}
\sum_i\|\alpha_i+\beta_i\|^2
&\le\sum_i(\|\alpha_i\|+\|\beta_i\|)^2\\
&=\sum_i\|\alpha_i\|^2+\sum_i\|\beta_i\|^2+2\sum_i\|\alpha_i\|\|\beta_i\|
\end{align}$$
Then we apply Cauchy-Schwarz inequality on the third term:
$$\begin{align}
\sum_i\|\alpha_i+\beta_i\|^2
&\le\sum_i\|\alpha_i\|^2+\sum_i\|\beta_i\|^2+2\sqrt{\sum_i\|\alpha_i\|^2}\sqrt{\sum_i\|\beta_i\|^2}\\
&=\left(\sqrt{\sum_i\|\alpha_i\|^2}+\sqrt{\sum_i\|\beta_i\|^2}\right)^2
\end{align}$$
The other side is similar:
$$\begin{align}
\sum_i\|\alpha_i+\beta_i\|^2
&\ge\sum_i(\|\alpha_i\|-\|\beta_i\|)^2\\
&=\sum_i\|\alpha_i\|^2+\sum_i\|\beta_i\|^2-2\sum_i\|\alpha_i\|\|\beta_i\|\\
&\ge\sum_i\|\alpha_i\|^2+\sum_i\|\beta_i\|^2-2\sqrt{\sum_i\|\alpha_i\|^2}\sqrt{\sum_i\|\beta_i\|^2}\\
&=\left(\sqrt{\sum_i\|\alpha_i\|^2}-\sqrt{\sum_i\|\beta_i\|^2}\right)^2
\end{align}$$
Now we are going to prove the main result:
Theorem 1
All quantum algorithms that find a solution with $O(1)$ probability requires $\Omega\left(\sqrt{\frac NM}\right)$ oracle queries.
Proof of Theorem 1
Without loss of generality, we may assume that the quantum algorithm uses $m$ qubits for some $m>n$. It applies $W$ unitary operations, interleaved with $W$ oracle operations. More specifically, let $|\psi\rangle$ be the initial state of the register, we compute $|\psi_W^R\rangle:=U_WO_RU_{W-1}O_R\cdots U_1O_R|\psi\rangle$, and then measure the first $n$ qubits of $|\psi_W^R\rangle$ as an answer. We may assume that the oracle queries are done on the first $n$ qubits, and the result is XORed with the $(n+1)$'th qubit. (Otherwise we can "swap" the queried qubits with the first $(n+1)$ qubits, and then swap them back after the query, as there is no limit on the unitary operations we apply.)
We define
$$\begin{align}
|\psi_k^R\rangle&:=U_kO_RU_{k-1}O_R\cdots U_1O_R|\psi\rangle\\
|\psi_k\rangle&:=U_kU_{k-1}\cdots U_1|\psi\rangle\\
D_k&:=\sum_{R\subset S}\left\|\psi_k^R-\psi_k\right\|^2
\end{align}$$
Upperbound of $D_W$
For the first half of the proof, we upperbound $D_k$ by $4k^2\binom{N-1}{M-1}$ using induction:
$$\begin{align}
D_{k+1}
&=\sum_{R\subset S}\left\|U_{k+1}O_R\psi_k^R-U_{k+1}\psi_k\right\|^2\\
&=\sum_{R\subset S}\left\|O_R\psi_k^R-\psi_k\right\|^2\\
&=\sum_{R\subset S}\left\|O_R(\psi_k^R-\psi_k)+(O_R-I)\psi_k\right\|^2\\
\end{align}$$
Notice that $O_R$ and $I$ can be written as
$$\begin{align}
O_R&=\sum_{x\notin R}|x\rangle\langle x|\otimes I\otimes I+\sum_{x\in R}|x\rangle\langle x|\otimes X\otimes I\\
I&=\sum_{x\notin R}|x\rangle\langle x|\otimes I\otimes I+\sum_{x\in R}|x\rangle\langle x|\otimes I\otimes I
\end{align}$$
Therefore we have
$$\begin{align}
D_{k+1}
&=\sum_{R\subset S}\left\|O_R(\psi_k^R-\psi_k)+\left(\sum_{x\in R}|x\rangle\langle x|\otimes(X-I)\otimes I\right)\psi_k\right\|^2
\end{align}$$
To apply Lemma 1, we upper bound the first term:
$$\begin{align}
\sum_{R\subset S}\left\|O_R(\psi_k^R-\psi_k)\right\|^2
&=\sum_{R\subset S}\left\|(\psi_k^R-\psi_k)\right\|^2\\
&=D_k\\
\end{align}$$
and the second term:
$$\begin{align}
&\sum_{R\subset S}\left\|\left(\sum_{x\in R}|x\rangle\langle x|\otimes(X-I)\otimes I\right)\psi_k\right\|^2\\
&=\sum_{R\subset S}\sum_{x\in R}\langle\psi_k|(|x\rangle\langle x|\otimes(2I-2X)\otimes I)|\psi_k\rangle\\
&=2\binom{N-1}{M-1}\sum_{x\in S}\langle\psi_k|(|x\rangle\langle x|\otimes(I-X))|\psi_k\rangle\\
&=2\binom{N-1}{M-1}(\langle\psi_k|\psi_k\rangle-\langle\psi_k|I\otimes X\otimes I|\psi_k\rangle)\\
&\le4\binom{N-1}{M-1}
\end{align}$$
By induction, $D_k\le4k^2\binom{N-1}{M-1}$, and with Lemma 1 we conclude that
$$\begin{align}
D_{k+1}
&\le\left(2k\sqrt{\binom{N-1}{M-1}}+2\sqrt{\binom{N-1}{M-1}}\right)^2\\
&=4(k+1)^2\binom{N-1}{M-1}
\end{align}$$
Lowerbound of $D_W$
For the second half of the proof, we lowerbound $D_k$ by $\Omega(1)\binom NM$. First, we define the projection matrix onto the subspace spanned by the solutions $R$:
$$\begin{align}
P_R:=\sum_{x\in R}|x\rangle\langle x|\otimes I
\end{align}$$
Then again, we split $D_W$ into two parts, and try to apply Lemma 1:
$$\begin{align}
D_W
&=\sum_{R\subset S}\left\|(I-P_R)\psi_W^R+(P_R\psi_W^R-\psi_W)\right\|^2\\
\end{align}$$
For the first term, we have $\langle\psi_W^R|(I-P_R)|\psi_W^R\rangle\le\frac12$, as we may assume that the probability of success is no less than $\frac12$ for this quantum algorithm. Thus we write down
$$\begin{align}
\sum_{R\subset S}\left\|(I-P_R)\psi_W^R\right\|^2\le\frac12\binom NM
\end{align}$$
For the second term, we have
$$\begin{align}
\left\|P_R\psi_W^R-\psi_W\right\|^2
&=1+\langle\psi_W^R|P_R|\psi_W^R\rangle-2\Re\langle\psi_W|P_R|\psi_W^R\rangle\\
&\ge\frac32-2\|P_R\psi_W\|^2\\
&=\frac32-2\langle\psi_W|P_R|\psi_W\rangle\\
\end{align}$$
Summing over $R$,
$$\begin{align}
\sum_{R\subset S}\left\|P_R\psi_W^R-\psi_W\right\|^2
&\ge\frac32\binom NM-2\sum_{R\subset S}\sum_{x\in R}\langle\psi_W|(|x\rangle\langle x|\otimes I)|\psi_W\rangle\\
&=\frac32\binom NM-2\binom{N-1}{M-1}\sum_{x\in S}\langle\psi_W|(|x\rangle\langle x|\otimes I)|\psi_W\rangle\\
&=\frac32\binom NM-2\binom{N-1}{M-1}\\
&=\left(\frac32-2\frac MN\right)\binom NM
\end{align}$$
We may assume that $\frac MN\le\frac14$. (In fact, for $\frac MN>\frac 14$, the problem is trivial as one only needs less than 4 trials on average using a naive algorithm.) We then apply Lemma 1:
$$\begin{align}
D_W
&\ge\left(\sqrt{\frac32-2\frac MN}-\sqrt{\frac12}\right)^2\binom NM\\
&\ge\left(\frac32-\sqrt2\right)\binom NM
\end{align}$$
Combining the lowerbound and upperbound of $D_W$, we have
$$\begin{align}
&\left(\frac32-\sqrt2\right)\binom NM\le D_W\le4W^2\binom{N-1}{M-1}\\
&\implies W\ge\frac{2-\sqrt2}4\sqrt{\frac NM}\\
&\implies W\ge\Omega\left(\sqrt{\frac NM}\right)
\end{align}$$
Which completes our proof.
