The answer is yes, with 2 distincts proofs for $q=p$ and $q\neq p$.

When $q=p$: Let $(x_n)_{n\ge 0}$ be generators of $C_{p^\infty}$ with $x_0=1$ and $x_{n+1}^p=x_n$ for all $n$. Define $y_n=x_n-1$ (in the group algebra $\mathbf{F}_p[C_{p^\infty}]$). Then $y_{n+1}^p=y_n$ for all $n$ and $y_0=0$. Define $I_n=y_nA$; then $I_n\subset I_{n+1}$ for all $n$; define $I_\infty=\bigcup I_n$. Then modulo $I_\infty$, every $x_n$ acts as the identity and hence every $\mathbf{F}_p$-subspace of $A$ containing $I_\infty$ is a submodule. Hence if $I_\infty\neq A$, then $A$ contains proper submodules of at most countable index (hence uncountable if $A$ is uncountable).

If $I_\infty=A$ and $A$ is uncountable, then $I_n$ is uncountable for some some $n$. If $I_n\neq A$, we are done. Otherwise, $y_nA=A$ and $y_n$ nilpotent force $A=0$, contradiction.

Now assume that $q\neq p$.

If $A$ is uncountable and splits as a direct sum of two submodules, one of the two is uncountable and we are done.

Assume otherwise ($A$ is "indecomposable"). Again denote by $x_n$ the generators of $C_{q^\infty}$ with $x_0=1$ and $x_{n+1}^q=x_n$. Then $x_n$ is an operator on $A$ vanished by some polynomial (namely $X^{q^n}-1$) with simple roots, hence for each $n$, we have a decomposition $A=\bigoplus_{P\in\mathcal{P}_n}A_P$, where $\mathcal{P}_n$ is the set of monic irreducible divisors of $X^{q^n}-1$ and $A_P=\mathrm{Ker}(P(x_n))$. Since $A$ is indecomposable, there exists a unique $P\in\mathcal{P}_n$ such that $A_P\neq 0$; denote it by $P_n$. So $P_n$ is irreducible and $P_n(x_n)=0$. Thus the subring $R_n$ of $\mathrm{End}_{\mathbf{F}_p}(A)$ generated by $x_n$ is a (finite) field. We have $R_n\subset R_{n+1}$ for all $n$ and hence $K=\bigcup R_n$ is a field as well; it is at most countable. On the other hand, the image of $C_{q^\infty}$ in $\mathrm{End}_{\mathbf{F}_p}(A)$ is contained in $K$. Hence any decomposition of $A$ as a $K$-module is $C_{q^\infty}$-invariant. In particular, we deduce that if $A$ is uncountable, then it is decomposable (as a direct sum of uncountably many countable submodules), contradiction.

The answer is no:

For all primes $p,q$, for every uncountable elementary abelian $p$-group, for every action of the Prüfer $q$-group $C_{q^\infty}\simeq\mathbf{Z}[1/q]/\mathbf{Z}$ on $A$ by group automorphisms, there exists a proper uncountable $C_{q^\infty}$-invariant subgroup in $A$.

There are 2 distincts proofs: for $q=p$ and $q\neq p$.

When $q=p$: Let $(x_n)_{n\ge 0}$ be generators of $C_{p^\infty}$ with $x_0=1$ and $x_{n+1}^p=x_n$ for all $n$. Define $y_n=x_n-1$ (in the group algebra $\mathbf{F}_p[C_{p^\infty}]$). Then $y_{n+1}^p=y_n$ for all $n$ and $y_0=0$. Define $I_n=y_nA$; then $I_n\subset I_{n+1}$ for all $n$; define $I_\infty=\bigcup I_n$. Then modulo $I_\infty$, every $x_n$ acts as the identity and hence every $\mathbf{F}_p$-subspace of $A$ containing $I_\infty$ is a submodule. Hence if $I_\infty\neq A$, then $A$ contains proper submodules of at most countable index (hence uncountable if $A$ is uncountable).

If $I_\infty=A$ and $A$ is uncountable, then $I_n$ is uncountable for some some $n$. If $I_n\neq A$, we are done. Otherwise, $y_nA=A$ and $y_n$ nilpotent force $A=0$, contradiction.

Now assume that $q\neq p$.

If $A$ is uncountable and splits as a direct sum of two submodules, one of the two is uncountable and we are done.

Assume otherwise ($A$ is "indecomposable"). Again denote by $x_n$ the generators of $C_{q^\infty}$ with $x_0=1$ and $x_{n+1}^q=x_n$. Then $x_n$ is an operator on $A$ vanished by some polynomial (namely $X^{q^n}-1$) with simple roots, hence for each $n$, we have a decomposition $A=\bigoplus_{P\in\mathcal{P}_n}A_P$, where $\mathcal{P}_n$ is the set of monic irreducible divisors of $X^{q^n}-1$ and $A_P=\mathrm{Ker}(P(x_n))$. Since $A$ is indecomposable, there exists a unique $P\in\mathcal{P}_n$ such that $A_P\neq 0$; denote it by $P_n$. So $P_n$ is irreducible and $P_n(x_n)=0$. Thus the subring $R_n$ of $\mathrm{End}_{\mathbf{F}_p}(A)$ generated by $x_n$ is a (finite) field. We have $R_n\subset R_{n+1}$ for all $n$ and hence $K=\bigcup R_n$ is a field as well; it is at most countable. On the other hand, the image of $C_{q^\infty}$ in $\mathrm{End}_{\mathbf{F}_p}(A)$ is contained in $K$. Hence any decomposition of $A$ as a $K$-module is $C_{q^\infty}$-invariant. In particular, we deduce that if $A$ is uncountable, then it is decomposable (as a direct sum of uncountably many countable submodules), contradiction.

Yes when $p=q$.

Let $(x_n)_{n\ge 0}$ be generators of $C_{p^\infty}$ with $x_0=1$ and $x_{n+1}^p=x_n$ for all $n$. Define $y_n=x_n-1$ (in the group algebra $\mathbf{F}_p[C_{p^\infty}]$). Then $y_{n+1}^p=y_n$ for all $n$ and $y_0=0$. Define $I_n=y_nA$; then $I_n\subset I_{n+1}$ for all $n$; define $I_\infty=\bigcup I_n$. Then modulo $I_\infty$, every $x_n$ acts as the identity and hence every $\mathbf{F}_p$-subspace of $A$ containing $I_\infty$ is a submodule. Hence if $I_\infty\neq A$, then $A$ contains proper submodules of at most countable index (hence uncountable if $A$ is uncountable).

If $I_\infty=A$ and $A$ is uncountable, then $I_n$ is uncountable for some some $n$. If $I_n\neq A$, we are done. Otherwise, $y_nA=A$ and $y_n$ nilpotent force $A=0$, contradiction.

The answer is yes, with 2 distincts proofs for $q=p$ and $q\neq p$.

When $q=p$: Let $(x_n)_{n\ge 0}$ be generators of $C_{p^\infty}$ with $x_0=1$ and $x_{n+1}^p=x_n$ for all $n$. Define $y_n=x_n-1$ (in the group algebra $\mathbf{F}_p[C_{p^\infty}]$). Then $y_{n+1}^p=y_n$ for all $n$ and $y_0=0$. Define $I_n=y_nA$; then $I_n\subset I_{n+1}$ for all $n$; define $I_\infty=\bigcup I_n$. Then modulo $I_\infty$, every $x_n$ acts as the identity and hence every $\mathbf{F}_p$-subspace of $A$ containing $I_\infty$ is a submodule. Hence if $I_\infty\neq A$, then $A$ contains proper submodules of at most countable index (hence uncountable if $A$ is uncountable).

If $I_\infty=A$ and $A$ is uncountable, then $I_n$ is uncountable for some some $n$. If $I_n\neq A$, we are done. Otherwise, $y_nA=A$ and $y_n$ nilpotent force $A=0$, contradiction.

Now assume that $q\neq p$.

If $A$ is uncountable and splits as a direct sum of two submodules, one of the two is uncountable and we are done.

Assume otherwise ($A$ is "indecomposable"). Again denote by $x_n$ the generators of $C_{q^\infty}$ with $x_0=1$ and $x_{n+1}^q=x_n$. Then $x_n$ is an operator on $A$ vanished by some polynomial (namely $X^{q^n}-1$) with simple roots, hence for each $n$, we have a decomposition $A=\bigoplus_{P\in\mathcal{P}_n}A_P$, where $\mathcal{P}_n$ is the set of monic irreducible divisors of $X^{q^n}-1$ and $A_P=\mathrm{Ker}(P(x_n))$. Since $A$ is indecomposable, there exists a unique $P\in\mathcal{P}_n$ such that $A_P\neq 0$; denote it by $P_n$. So $P_n$ is irreducible and $P_n(x_n)=0$. Thus the subring $R_n$ of $\mathrm{End}_{\mathbf{F}_p}(A)$ generated by $x_n$ is a (finite) field. We have $R_n\subset R_{n+1}$ for all $n$ and hence $K=\bigcup R_n$ is a field as well; it is at most countable. On the other hand, the image of $C_{q^\infty}$ in $\mathrm{End}_{\mathbf{F}_p}(A)$ is contained in $K$. Hence any decomposition of $A$ as a $K$-module is $C_{q^\infty}$-invariant. In particular, we deduce that if $A$ is uncountable, then it is decomposable (as a direct sum of uncountably many countable submodules), contradiction.