No, the different $p$-adic number systems are not in any way compatible with one another.
A $p$-adic number is a not a number that is $p$-adic; it is a $p$-adic number. Similarly, a real number is not a number that is real, it is a real number. There is not some unified notion of "number" that these are all subsets of; they are entirely separate things, though there may be ways of identifying bits of them in some cases (e.g., all of them contain a copy of the rational numbers).
Now, someone here is bound to point out that if we take the algebraic closure of some $\mathbb{Q}_p$, the result will be algebraically isomorphic to $\mathbb{C}$. But when we talk about $p$-adic numbers we are not just talking about their algebra, but also their absolute value, or at least their topology; and once you account for this they are truly different. (And even if you just want algebraic isomorphism, this requires the axiom of choice; you can't actually identify a specific isomorphism, and there's certainly not any natural way to do so.)
How can we see that they are truly different? Well, first let's look at the algebra. The $5$-adics, for instance, contain a square root of $-1$, while the $3$-adics do not. So if you write down a $5$-adic number which squares to $-1$, there cannot be any corresponding $3$-adic number.
But above I claimed something stronger -- that once you account for the topology, there is no way to piece the various $p$-adic number systems together, which the above does not rule out. How can we see this? Well, let's look at the topology when we look at the rational numbers, the various $p$-adic topologies on $\mathbb{Q}$. These topologies are not only distinct -- any finite set of them is independent, meaning that if we let $\mathbb{Q}_i$ be $\mathbb{Q}$ with the $i$'th topology we're considering, then the diagonal is dense in $\mathbb{Q}_1 \times \ldots \times \mathbb{Q}_n$.
Put another way -- since these topologies all come from metrics -- this means that for any $c_1,\ldots,c_n\in\mathbb{Q}$, there exists a sequence of rational numbers $a_1,a_2,\ldots$ such that in topology number 1, this converges to $c_1$, but in topology number two, it converges to $c_2$, and so forth. (In fact, more generally, given any finite set of inequivalent absolute values on a field, the resulting topologies will be independent.)
So even on $\mathbb{Q}$, the different topologies utterly fail to match up, so there is no way they can be pieced together by passing to some larger setting.
Because is says that if $x=\cdots1313_5$ then $3x+1\equiv 0 \text{ mod } 5^n$ for all $n$, which is precisely what it means to be $0$ in $\mathbb{Q}_5$. Thus, you see that $3x+1=0$ so that $\displaystyle x=\frac{-1}{3}$.
EDIT:
Now that I have more time, let me be less glib about this response.
Whenever possible, we want to turn problems about $\def\Qp{\mathbb{Q}_5}$ $\def\Zp{\mathbb{Z}_5}$ $\Qp$ into problems about $\Zp$ since they are easy to deal with. So, how can we interpret $\displaystyle \frac{-1}{3}\in\Qp$, well since $\Qp$ is $\text{Frac}(\Zp)$ the only clear interpretation is that it is the element $x$ of $\Zp$ which satisfies $3x+1=0$. So, instead let us try to look for a solution $3x+1=0$.
To begin, let us recall how we can think about $\Zp$. Intuitively, $\Zp$ is the set $\{z\}$ of solutions to systems of equations as follows:
$$\begin{cases}z &\equiv a_1 \mod 5\\ z &\equiv a_2 \mod 5^2\\ z &\equiv a_3 \mod 5^3\\ &\vdots\end{cases}$$
where the equations are "consistent" (i.e. $a_i\equiv a_j\mod p^i$ for $i\leqslant j$). So, now if $x$ satisfies $3x+1=0$ then this should translate to mean
$$\begin{cases}3x+1 &\equiv 3a_1+1 \equiv 0 \mod 5\\ 3x+1 &\equiv 3a_2+1 \equiv 0 \mod 5^2\\ 3x+1 &\equiv 3a_3+1\equiv 0 \mod 5^3\\ &\vdots\end{cases}$$
So, we can solve each of these equations piecewise and find that
$$(a_1,a_2,a_3,\ldots)=(3,8,83,\ldots)$$
Ok, so, this tells us that $x=(3,8,83,\ldots)$...this doesn't look right? How can we go from this to the desired $x=\ldots1313_5$? The key is that we have the same element of $\Qp$ expressed in different forms. Indeed, the notation $x=1313_5$ means that
$$x=3+1\cdot 5+3\cdot 5^2 +\cdots$$
To reconcile this ostensible difference, let us write $3+1\cdot 5+3\cdot 5^2+\cdots$ in the same notation that we already have $x$ in. Recall that the correspondence between $\mathbb{Z}$ and these sequences is
$$m\mapsto (m\mod 5,m\mod 5^2,m\mod 5^3,\ldots)$$
Thus, we see that
$$\begin{aligned} 3 & \mapsto (3,3,3,\cdots)\\ 5 &\mapsto (0,5,5,\ldots)\\ 5^2 & \mapsto (0,0,5^2,\ldots)\end{aligned}$$
Thus, we see that
$$ \begin{aligned}3+5+3\cdot 5^2 &=(3,3,3,\ldots)+(0,5,5,\ldots)+(0,0,75,\ldots)\\ &= (3,8,83,\ldots)\end{aligned}$$
and voilà!
Best Answer
Although there may be special arguments in special situations, as far as I know, the only way to find the $p$-adic expansion of a rational number is by computation, either with computer help or by hand. Here’s a method.
First, I recommend using $p$-ary notation, so that $\dots dcba;_p$ represents $a+bp+cp^2+dp^3+\cdots$, just as when you write a positive integer in $p$-ary. I’ll do an example of $5$-adic division in $5$-ary notation. Let’s expand $\frac78$ as an element of $\Bbb Z_5$. Unfortunately, the method I like best can be explained very nicely in a lecture, but is almost impossible to explain in a static medium like print. I’ll show you a different approach that’s somewhat less general. First use convergent geometric series to write out the expansion of $1/(1-5)=-\frac14=1+5+5^2+5^3+\cdots=\cdots11111;_5$, and notice that this means that $-1=\cdots44444;_5$. Now you see how to change the sign of a $5$-adic number in this notation. When you subtract from $-1$, you get an exchange of digits, $0\leftrightarrow4$, $1\leftrightarrow3$, and $2\leftrightarrow2$. Then to get $-z$, you first form $-1-z$, and then add $1$ to the result.
But continuing the expansion via convergent geometric series, we have $1/(1-5^2)=-\frac1{24}=1+5^2+5^4+5^6+\cdots=\cdots010101;_5$, and $-\frac1{124}=\cdots1001001001;_5$, $-\frac1{624}=\cdots100010001;_5$, etc. What about $-\frac18$? That’s $3\cdot(-\frac1{24})=\cdots3030303;_5$, and since $7=12;_5$ and $3$ times this is $41;_5$, we get $-\frac78=\cdots414141;_5$. When we reverse the digits we get $-1-(-\frac78)=\cdots03030303;_5$ then add $1$ to get $\frac78=\cdots03030304;_5\,$.