If by $f_1\equiv f_2$ modulo $p$, you mean that $a_n(f_1)\equiv a_{n}(f_2)$ modulo $p$ for all $n\in\mathbb N$ or maybe for all except finitely many, then this theorem cannot be true.
Let's start with a counterexample. Consider the two newforms
\begin{equation}
f(z)=\underset{n=1}{\overset{\infty}{\Sigma}}a_{n}q^{n}\in S_{2}(\Gamma_0(11)),\ g(z)=\underset{n=1}{\overset{\infty}{\Sigma}}b_{n}q^{n}\in S_{2}(\Gamma_0(77))
\end{equation}
attached respectively to the elliptic curves $X_0(11)$ and $E:y^2+xy=x^3+x^2-51x+110$. Then $a_{\ell}(f)\equiv a_{\ell}(g)$ modulo 3 for all primes $\ell\neq 7$ yet $a_7(f)=-2$ while $a_7(g)=-1$ so $a_7(f)\not\equiv a_7(g)$ modulo 3, and in fact typically $a_n(f)\not\equiv a_n(g)$ modulo 3 if $n$ is a strictly positive integer which is divisible by 7.
What's going on? The problem is that $a_\ell$ is the trace of the image of the Frobenius morphism at $\ell$ through the Galois representation attached to the newform provided $\ell$ does not divide the conductor of the newform. The fact that $a_{\ell}(f)\equiv a_\ell(g)$ for almost all primes implies that the residual representations $\bar{\rho}_f$ and $\bar{\rho}_g$ are isomorphic (by Cebotarev's density theorem). If $\ell$ divides neither the conductor of $f$ nor the conductor of $g$, then $a_\ell(f)$ modulo $p$ and $a_\ell(g)$ modulo $p$ are both the trace of the image of the Frobenius morphism, so they are equal. However, 7 divides the conductor of $g$ but not the Artin conductor of the residual representation $\bar{\rho}_g$ nor the conductor of $f$ so the relation between $a_7(g)$ and $a_7(f)\equiv\mathrm{tr}\bar{\rho}_g(\mathrm{Fr}(7))$ breaks down.
This will happen generally speaking whenever you have a level-raising (or equivalently level-lowering) phenomenon, that is to say congruences between forms which are new of different levels.