Why insist on dyadic decomposition of cubes when you don't need it?
Observe that the maximal function $\mathcal{M}f = \mathcal{M} |f|$ by definition. And observe that since $\phi$ is positive, $|\phi_t\ast f| \leq \phi_t \ast |f|$. Hence we can assume without loss of generality that $f$ is positive.
Let $\lambda_\phi(s)$ for $s \geq 0$ be the set $\{ \phi(x) \geq s\}$. We have that $\phi(x) = \int_0^P \chi_{\lambda_{\phi}(s)}(x) \mathrm{d}s$, where $P = \sup \phi$. Note that by assumption $\lambda_\phi(s)$ for a decreasing family of balls around the origin. Let $\lambda^*\phi(s)$ be the ball whose radius is 1 more than the radius of $\lambda_\phi(s)$. Now, let $|y-x| < 1$. We have that
$$ \int_{\lambda_\phi(s)} f(y-z) \mathrm{d}z \leq \int_{\lambda^*_\phi(s)} f(x - z) \mathrm{d}z $$
since $f$ is positive and $x + \lambda^*_\phi(s) \supset y + \lambda_\phi(s)$.
Let $P'$ be the smallest number such that $\lambda_\phi(P')$ has radius at most 1. By assumption (that $\phi$ is a bounded integrable function) we have that $\lambda_\phi(P')$ has positive radius $R'$.
$$\begin{align} \phi\ast f(y) &= \int_0^{P'} \int_{\lambda_\phi(s)} f(y-z) \mathrm{d}z \mathrm{d}s + \int_{P'}^P \int_{\lambda_\phi(s)} f(y-z) \mathrm{d}z \mathrm{d}s \\
&\leq \int_0^{P'} \int_{\lambda^*_\phi(s)} f(x-z) \mathrm{d}z \mathrm{d}s + \int_{P'}^P\int_{\lambda^*_\phi(P')} f(x-z) \mathrm{d}z \mathrm{d}s \\
& \leq \int_0^{P'} |\lambda_\phi^*(s)| \mathcal{M}f(x) \mathrm{d}s + |P' - P||\lambda_\phi^*(P')|\mathcal{M}f(x)
\end{align}$$
Now using that for $s \leq P'$ we have that $|\lambda_\phi^*(s)| \leq c^n |\lambda_\phi(s)|$ where $c = 1 + 1/R'$ we have
$$ \leq \mathcal{M}f(x)\left( c^n \int_0^P |\lambda_\phi(s)| \mathrm{d}s + |P-P'| |\lambda_\phi^*(P')|\right)\tag{*}$$
The first term inside the parenthesis gives $\int \phi(z) \mathrm{d}z$ from the distributional function characterisation of Lebesgue spaces. The second term is quite obviously a finite constant depending on $\phi$.
Now, if we replace $\phi$ by $\phi_t$ in the above argument, then $P \mapsto t^{-n} P$. We will need to consider $|y-x| < t$ and we let $\lambda^*_{\phi_t}(s)$ to have radius $t$ more than its counterpart without the star. We also let $P'$ be such that the corresponding ball has radius at least $t$: by the scaling property of $\phi_t$ we see that $P' \mapsto t^{-n} P'$, and $\lambda_\phi(P') \to t \lambda_\phi(P')$. So the above analysis goes to show that the constant derived above (the term inside the parentheses in (*)) does not depend on the scaling $t$. Hence we get the desired inequality.
Recall that $f:\Bbb{R}^d\rightarrow \Bbb{R}$ is Lebesgue measurable if $\{f>\alpha\}$ is open for every real number $\alpha$ (this follows from the standard definition that $f$ is measurable if $f^{-1}([-\infty,\alpha))$ is measurable).
Then, let the maximal function be defined as usual
$$
Mf(x)=\sup_{B\ni x}\frac{1}{\vert B\vert}\int_B\vert f(y)\vert dy
$$
Now, $\{Mf>\alpha\}$ is open since if $y\in \{Mf>\alpha\}$, there is a ball $B$ such that $y\in B$ and
$$
\frac{1}{\vert B\vert}\int_B\vert f\vert >\alpha
$$ And, for any other $x\in B$, we have
$$
Mf(x)\geq\frac{1}{\vert B\vert}\int_B\vert f\vert>\alpha
$$and hence $x\in \{Mf>\alpha\}$ as well.
Best Answer
No. By Lebesgue differentiation theorem, $g=f$ almost everywhere. Now let $f(x)=x^2 \mathbf 1_{[-10,10]}\in L^1$. $f\ge 0$. Then, for all $x\in[-9,9]$, a lower bound for $f^*$ is given by $\frac12\int_{x-1}^{x+1} y^2 dy = x^2 + \frac13$. In particular, $f^*>f=g$, at least near $x=0$. Here's a Desmos graph to verify. In fact $f^*(0)=100/3$, which is much larger than $f(0)=0$. You can similarly show that $f^*(x)>f(x)$ for all $x\in(-10,10)$.
In addition, for any $x$ with $|x|>10$, there will always be some (possibly large) ball around $x$ that intersects $[-10,10]$. This implies $f^*(x)>0=f(x)$. In conclusion, $f^*>f$ almost everywhere.
One can make this last part for $|x|>10$ more precise (and this is close to what you might read in a book)： the ball of radius $|x|+10$ around $x$ completely contains $[-10,10]$. Consequently, $$f^*(x)>\frac1{2|x|+20}\int_{-10}^{10} y^2 dy \ge \frac{C}{1+|x|}.$$ This proves that $f^*$ is not integrable at infinity.