Metric signature conventions: minus sign for $x^a$ or $x_a$? Say I use the metric signature $(-+++)$. Then
$\partial_a=(\partial_0,\partial_i)=(-\partial^0,\partial^i)$,
but
$\partial^a=(\partial^0,\partial^i)=(-\partial_0,\partial_i)$.
The same goes for $p^a$ and $p_a$, I suppose. I know that we need to contract, say, $x^a$ with $p_a$ to give $-x^0p_0+x^ip_i$. So my question is: Does this convention of whether to put ``$-$'' on quantities with superscripts or subscripts apply to all quantities? $x^a$ and $x_a$? $A^a$ and $A_a$?
 A: In general, when replacing a free index with a specific one, no signs ever get introduced:
\begin{align}
\partial_a & \to (\partial_0, \partial_i) \\
\partial^a & \to (\partial^0, \partial^i).
\end{align}
This holds for all signatures. (On a side note, I'm being pedantic about not using "$=$" signs for a reason - a tensor is not equal to a single, albeit unspecified, component of itself.)
The only question, then, is the relation between $\partial_0$ and $\partial^0$, and between $\partial_i$ and $\partial^i$. In general, again without regard to signature, we have
\begin{align}
\partial_a & = g_{ab} \partial^b \\
\partial^a & = g^{ab} \partial_b.
\end{align}
In special relativity, $g_{ab} = \eta_{ab}$ and $g^{ab} = \eta^{ab}$, where now we need to agree on a signature to conclude
\begin{align}
\partial_0 & = -\partial^0 & \partial_i & = \partial^i \\
\partial^0 & = -\partial_0 & \partial^i & = \partial_i.
\end{align}
Actually, your worry about where to put the signs means you've put in an extra. In fact, we have
$$ x^a p_a = x^0 p_0 + x^1 p_1 + x^2 p_2 + x^3 p_3. $$
This is exactly the same as
$$ x_a p^a = x_0 p^0 + x_1 p^1 + x_2 p^2 + x_3 p^3. $$
The above holds in GR in any signature. If you are in SR and have the $(-,+,+,+)$ signature, then you can write
$$ x^a p_a = -x^0 p^0 + x^1 p^1 + x^2 p^2 + x^3 p^3 $$
or
$$ x^a p_a = -x_0 p_0 + x_1 p_1 + x_2 p_2 + x_3 p_3. $$
Note the negative term only enters when you insist on using the same indices (either upper or lower) for both $x$ and $p$. Also note that these two formulas only hold in SR, where swapping indices at worst introduced a negative. In general,
$$ x^a p_a = g_{ab} x^a p^b, $$
which can have $16$ terms in the sum.
A: Using your metric signature, the metric is:
$$\eta_{\mu\nu}=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$
The generic position vector is defined as (in Cartesian) $$x^{\mu}=\begin{pmatrix}t\\x\\y\\z\end{pmatrix}$$
And the quantity $x_{\mu}=\eta_{\mu\nu}x^{\nu}$. So you can see that $x_\mu$ becomes the one with a minus sign on $x_0$.
Then, the derivative is defined as:
$$\partial_\mu=\frac{\partial}{\partial x^\mu}$$
So you can see that $\partial_0=\frac{\partial}{\partial t}$ is positive. When you apply the metric, $$\partial^\mu=\eta^{\mu\nu}\partial_\nu=\eta^{\mu\nu}\frac{\partial}{\partial x^\nu}=\frac{\partial}{\partial x_\mu}$$
and thus $\partial^0$ becomes negative.
So to answer your question, yes the convention applies. Changing a superscript to a subscript is done through applying the metric. $A_a=\eta_{ab}A^b$ and $A^a=\eta^{ab}A_b$ and $\eta^{ab}\eta_{ab}=4$.
A: Yes, indeed it does.
$\\$
Have you taken a maths course on Metric Spaces perhaps?
A Metric Space, roughly, is something where we have a notion of how to measure things.
So, for example, Euclidean Space is a very fimiliar space, you measure the square of the magnitude of a vector to be
$$ \vec{x} \cdot \vec{x} = (x^1)^2 + (x^2)^2 + (x^3)^2 = (x_1)^2 + (x_2)^2 + (x_3)^2 $$
This could also be written using a metric as 
$$ \vec{x} \cdot \vec{x} = g_{ij} x^i x^j $$
where $i = 1,2,3$ and $j = 1,2,3$. Here I'm using Einstein Summation Convention, such that $g_{ij} x^i x^j$ with the $i$ and $j$ repeated like that, one upper and one lower, is shorthand for
$$ g_{ij} x^i x^j = \sum_{i=1}^3 \sum_{j=1}^3 g_{ij} x^i x^j$$
$\\$
So our metric, $g_{ij}$ is just 
$$ g_{ij} = \mbox{diag}(+1,+1,+1) $$
that is, when $i=j$ we get $g_{ii} = + 1$, while if $i \neq j$ we get zero, $g_{ij}\vert_{i\neq j} =0$.
$\\$
You'll then see that when raising and lowering indices using the metric like
$$ x^i = g^{ij} x_j \quad \mbox{ or } \quad x_i = g_{ij} x^j $$
That the raised one $x^i$ and the lowered one $x_j$ are the same, since the $g_{ij}$ are each just $+1$.
So that was nice and easy.
$\\$
Now when we move to Minkowski Space for Special Relativity, it gets a little more complicated. The metric components (the equivalent of the $g_{ij}$) are no longer all $+1$. We usually use $\eta^{\mu \nu}$ for the Minkowski Space metric, where
$$ \eta^{\mu \nu} = \mbox{diag}(-1,+1,+1,+1)  $$
in your convention. (It should be noted that not everyone uses the same set of +1 and -1, so you should always check this and make sure. It can be very fustrating when something doesn't work out because the other guy was using a different sign!).
Anyway, a vector, like $A^{\mu}$ as you ask, can be lowered with the metric tensor like
$$ A_{\mu} = g_{\mu \nu} A^{\nu} $$
So since by definition 
$$ A_{\mu} = (A^0, A^i) $$
when we lower the index we get
$$ A_{\mu} = g_{\mu \nu} A^{\nu} = (g_{0 \nu} A^{\nu}, g_{i \nu} A^{\nu}) = \left( \sum_{\nu = 0}^3 g_{0 \nu} A^{\nu}, \sum_{\nu = 0}^3 g_{i \nu} A^{\nu} \right)  $$
But since all the $g_{\mu \nu}$ are zero when $\mu \neq \nu$, the only ones in the sum that are not zero are
$$ A_{\mu} = g_{\mu \nu} A^{\nu} = (g_{0 0} A^{0}, g_{i i} A^{i})  $$
and we know this from the definition of our metric tensor above, namely
$$ A_{\mu} = g_{\mu \nu} A^{\nu} = ((-1) A^{0}, (+1) A^{i}) = (-A^{0}, A^{i}) $$
$\\$
You can use the metric tensor to raise and lower any index, such as, for example,  
$$ T_{\mu \nu \rho}^{\; \; \; \; \sigma} = g_{\mu \alpha} g_{\nu \beta} g^{\sigma \gamma} T^{\alpha \beta}_{\; \; \; \rho \gamma} $$
