For $n\le 3$ this is certainly true if $p \ge 5$ is assumed, and this is directly follows from work of Ljunggren and Jacobsthal discussed later. 

This should usually fail for $n \ge 4$. I give below an informal explanation that suggests that given any $a,b$, your congruence fails if $n$ is large enough with respect to $a$ and $b$. Concretely, $n > 3+\nu_p(\binom{a}{b}ab(a-b))$ should work (or rather, fail), where $\nu_p$ is the $p$-adic valuation.

For example, let us take $a=2$, $b=1$ and $n=4$. You want
$$\binom{2p^4}{p^4} \equiv 2 \bmod p^4.$$
Here is why it fails. By Ljunggren--Jacobsthal,
$$\binom{2p^4}{p^4} \equiv \binom{2p^3}{p^3} \bmod p^{12},$$
$$\binom{2p^3}{p^3} \equiv \binom{2p^2}{p^2} \bmod p^{9},$$
$$\binom{2p^2}{p^2} \equiv \binom{2p}{p} \bmod p^{6}.$$
In particular, we do know that 
$$\binom{2p^4}{p^4} \equiv \binom{2p}{p} \bmod p^4.$$
You congruence then implies
$$\binom{2p}{p} \bmod 2 p^4.$$
Primes that satisfy this are called Wolstenholme primes. The only such primes up to $10^9$ are 16843 and 2124679.

After the informal explanation I give an overview of a modified congruence which does hold.

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Heuristic for failure: I approach this using $p$-adic analysis as explained in Chapter 7 of Alain M. Robert's book "A course in $p$-adic analysis". For $p \ge 3$, let $\Gamma_p$ be the $p$-adic Gamma function, which is a continuous function on $\mathbb{Z}_p$, whose value at positive integers is
$$ \Gamma_p(n) =(-1)^n \prod_{\substack{1 \le i <n\\ p \nmid i}}i. $$
It is straightforward to verify
$$ \binom{ap^n}{bp^n} = \frac{\Gamma_p(ap^n)}{\Gamma_p(bp^n)\Gamma_p((a-b)p^n)} \binom{ap^{n-1}}{bp^{n-1}}.$$
Iterating this identity, we find
$$ \binom{ap^n}{bp^n} =  \prod_{i=1}^{n} \frac{\Gamma_p(ap^i)}{\Gamma_p(bp^i)\Gamma_p((a-b)p^i)} \binom{a}{b}.$$
If one wants to establish your congruence, it suffices to show that
$$ \prod_{0=1}^{n-1} \frac{\Gamma_p(ap^i)}{\Gamma_p(bp^i)\Gamma_p((a-b)p^i)} \equiv 1 \bmod p^n.$$
(In fact, replacing the exponent $n$ with $n$ minus the $p$-adic valuation of $\binom{a}{b}$, this is in fact equivalent to your claim.)
Let 
$$f(x):=\log \Gamma_p(px)$$
as in page 381 of Robert's book. It has a Taylor expansion with coefficients in $p \mathbb{Z}_p$.
Using the isometry property of the $p$-adic logarithm (see page 380), we can formulate our problem as establishing
$$ \left| \sum_{i=0}^{n-1} (f(ap^{i}) - f(bp^i)-f((b-a)p^i))\right| \le p^{-n}.$$
We now use the expansion
$$ f(x)=\lambda_0 px - \sum_{j \ge 1}\frac{\lambda_j}{2j(2j+1)}p^{2J+1}x^{2j+1}$$
where $\lambda_0 \in \mathbb{Z}_p$ and $p\lambda_j \in \mathbb{Z}_p$ for $j \ge 1$. (See page 381.) In fact, $\lambda_1 \in \mathbb{Z}_p$ as shown in the paper of Robert--Zuber mentioned below.

We have
$$ f(x+y)-f(x)-f(y) = -\sum_{j \ge 1} \frac{\lambda_j}{2j(2j+1)} p^{2j+1} ((x+y)^{2j+1}-x^{2j+1}-y^{2j+1})$$
since the linear term vanishes, and
$$ f(ap^i)-f(bp^i)-f((a-b)p^i) = -\sum_{j \ge 1} \frac{\lambda_j}{2j(2j+1)}  p^{(i+1)(2j+1)} (a^{2j+1}-b^{2j+1}-(a-b)^{2j+1})$$
so we want 
$$\left| \sum_{j \ge 1} \frac{\lambda_j}{2j(2j+1)}  (a^{2j+1}-b^{2j+1}-(a-b)^{2j+1})\sum_{i=0}^{n-1} p^{(i+1)(2j+1)} \right|\le p^{-n}.$$
One might hope that your congruence is explained by each $j$-term being $\le p^{-n}$. The first term is $p^3$ times
$$\frac{p^{3n}-1}{p^3-1} \frac{\lambda_1}{2} 3ab(a-b).$$
The only part that depends on $n$ is $(p^{3n}-1)/(p^3-1)$, which has $p$-adic valuation equal to $0$. Hence, I expect that for fixed $a,b$ and large enough $n$ you will always get a counterexample.

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The congruence I am familiar with is
$$ \binom{p^k a}{p^k b} \equiv \binom{p^{k-1} a}{p^{k-1}b} \bmod p^{3k}$$
for $p\ge 5$.

This is originally due to Ljunggren and Jacobsthal in "On the divisibility of the difference between two binomial coefficients" in Skand. Mat.-Kongr., Trondheim 1949, 42-54 (1952). The case $k=1$ was established in the 19th century by Wolstenholme, with earlier works (with smaller exponents) by Cabbage and Lucas.

It is also attributed to G. S. Kazandzidis, who proved a somewhat stronger result. See his papers

 - "On a congruence and on a practical method for finding the highest power of a prime p which divides the binomial coefficient (AB)", Bull. Soc. Math. Grèce, N. Sér. 6, No. 2, 358-360 (1965).
 - "Congruences on the binomial coefficients", Bull. Soc. Math. Grèce, N. Ser. 9, No. 1, 1-12 (1968).
 - "On congruences in number-theory", Bull. Soc. Math. Grèce, N. Sér. 10, No. 1, 35-40 (1969).

Kazandzidis proved $$ \binom{pn}{pk} \equiv \binom{n}{k} \bmod p^{3+a}$$
where $a$ is the $p$-adic valuation of $nk(n-k)\binom{n}{k}$.

All these proofs are elementary. The modern treatment is a bit less so, but is more illuminating. Indeed, $p$-adic analysis can lead to a proof. See Chapter 7 of Alain M. Robert's book "A course in p-adic analysis" (GTM 198, Springer, 2000). The proof given there is based on the material in the paper "The Kazandzidis supercongruences. A simple proof and an application", Rend. Semin. Mat. Univ. Padova 94, 235-243 (1995), by Alain M. Robert and Maxime Zuber.

Some remarks:

 - The congruence $\binom{ap^k}{bp^k} \equiv \binom{ap^{k-1}}{bp^{k-1}} \bmod p^{2k}$ has a combinatorial proof (as opposed to algebraic proofs mod $p^{3n}$). A reference for a combinatorial proof when the modulus is $p^n$ is Gian-Carlo Rota and Bruce Sagan's paper "Congruences derived from group action", Eur. J. Comb. 1, 67-76 (1980). I am not sure if this is the earliest reference of this kind of argument, though.
 - There are various $q$-analogues of this $p$-adic congruences.
 - The above papers include also the cases $p=3$ (and maybe $p=2$?) for which the exponent of $p$ in the modulus is slightly smaller.